ACM算法模板 · 一些常用的算法模板-模板合集（稍加修改）

 原博客 https://blog.youkuaiyun.com/qq_32265245/article/details/53046750

 

附： 星期数计算模板

int week(int n,int y,int r)
{
    //用来特判闰年
    if( y ==1 || y==2)
    {
        y = y+12;
        n --;
    }
    int ans = (r+2*y+3*(y+1)/5+n+n/4-n/100+n/400)%7;
    return ans+1;
}

 

0.头文件

 

#define _CRT_SECURE_NO_DEPRECATE
#include <set>  
#include <cmath>  
#include <queue>  
#include <stack>  
#include <vector>  
#include <string>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
#include <functional>  
#include<string.h>  
#include<stdio.h>  
using namespace std;  
  
const int maxn = 110;  
const int INF = 0x3f3f3f3f;  

 

 

 

经典

1.埃拉托斯特尼筛法(求素数)

/*
    |埃式筛法|
    |快速筛选素数|
    |16/11/05ztx|
*/
int qwq[10005]; //这个数组用来存找到的素数
int cnt=0;
int prime[maxn];
bool is_prime[maxn];

int sieve(int n){
    int p = 0;
    for(int i = 0; i <= n; ++i)
        is_prime[i] = true;
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i <= n; ++i){   //  注意数组大小是n
        if(is_prime[i]){
            prime[p++] = i;
            qwq[cnt]=i; cnt++;  //开始存
            for(int j = i + i; j <= n; j += i)  //  轻剪枝，j必定是i的倍数
                is_prime[j] = false;
        }
    }
    return p;   //  返回素数个数
}


int main()
{
    cout<<sieve(100)<<endl;  //输出100内的素数个数，紧接着输出
    for(int i=0;i<cnt;i++) cout<<qwq[i]<<" ";
}

 

 

2.快速幂

 

 

/*
    |快速幂| 传入的分别是x的n次方 并取余m
*/


long long  qwq(long long x, long long  n, long long m)
{
    long long res = 1;
    while (n > 0){
        if  (n & 1) //  判断是否为奇数，若是则true  
            res = (res * x) % m;
        x = (x * x) % m;
        n >>= 1;    //  相当于n /= 2; 
    }
    return res;
}

 

3.大数模拟

 

大数加法

/*
    |大数模拟加法|
    |用string模拟|
    |16/11/05ztx, thanks to caojiji|
*/

string add1(string s1, string s2)
{
    if (s1 == "" && s2 == "")   return "0";
    if (s1 == "")   return s2;
    if (s2 == "")   return s1;
    string maxx = s1, minn = s2;
    if (s1.length() < s2.length()){
        maxx = s2;
        minn = s1;
    }
    int a = maxx.length() - 1, b = minn.length() - 1;
    for (int i = b; i >= 0; --i){
        maxx[a--] += minn[i] - '0'; //  a一直在减 ， 额外还要减个'0'
    }
    for (int i = maxx.length()-1; i > 0;--i){
        if (maxx[i] > '9'){
            maxx[i] -= 10;//注意这个是减10
            maxx[i - 1]++;
        }
    }
    if (maxx[0] > '9'){
        maxx[0] -= 10;
        maxx = '1' + maxx;
    }
    return maxx;
}

 

大数阶乘

 

/*
    |大数模拟阶乘|
    |用数组模拟|
    |16/12/02ztx|
*/

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long LL;

const int maxn = 100010;

int num[maxn], len;

/*
    在mult函数中，形参部分：len每次调用函数都会发生改变，n表示每次要乘以的数，最终返回的是结果的长度
    tip: 阶乘都是先求之前的(n-1)!来求n!
    初始化Init函数很重要，不要落下
*/

void Init() {
    len = 1;
    num[0] = 1;
}

int mult(int num[], int len, int n) {
    LL tmp = 0;
    for(LL i = 0; i < len; ++i) {
         tmp = tmp + num[i] * n;    //从最低位开始，等号左边的tmp表示当前位，右边的tmp表示进位（之前进的位）
         num[i] = tmp % 10; //  保存在对应的数组位置，即去掉进位后的一位数
         tmp = tmp / 10;    //  取整用于再次循环,与n和下一个位置的乘积相加
    }
    while(tmp) {    //  之后的进位处理
         num[len++] = tmp % 10;
         tmp = tmp / 10;
    }
    return len;
}

int main() {
    Init();
    int n;
    n = 1977; // 求的阶乘数
    for(int i = 2; i <= n; ++i) {
        len = mult(num, len, i);
    }
    for(int i = len - 1; i >= 0; --i)
        printf("%d",num[i]);    //  从最高位依次输出,数据比较多采用printf输出
    printf("\n");
    return 0;
}

 

4.GCD(求最大公约数)

 

/*
    |辗转相除法|
    |欧几里得算法|
    |求最大公约数|
    |16/11/05ztx|
*/

int gcd(int big, int small)
{
    if (small > big) swap(big, small);
    int temp;
    while (small != 0){ //  辗转相除法
        if (small > big) swap(big, small);
        temp = big % small;
        big = small;
        small = temp;
    }
    return(big);
}

 

5.LCM（求最小公倍数）

 

int n,m;
int LCM(int big, int small)
{
    if (small > big) swap(big, small);
    int temp;
    while (small != 0){ //  辗转相除法
        if (small > big) swap(big, small);
        temp = big % small;
        big = small;
        small = temp;
    }
    return(n*m/big);
}
int main()
{
    cin>>n>>m;
    cout<<LCM(n,m);
}

 

 

6.全排列

 

 

#include<bits/stdc++.h>
using namespace std;

//k是开始的元素标号，n是开始的元素标号+要排列的元素数
//这个规则和sort类似
//若只对a[1]到a[4]进行全排列 则k=1，n=5；
//排列a[0]到a[4],则k=0，n=5
void Pern(int list[], int k, int n)
{   //  k表示前k个数不动仅移动后面n-k位数
    if (k == n - 1) {
        for (int i = 0; i < n; i++) {
            printf("%d", list[i]);
        }
        printf("\n");
    }else {
        for (int i = k; i < n; i++) {   //  输出的是满足移动条件所有全排列
            swap(list[k], list[i]);
            Pern(list, k + 1, n);
            swap(list[k], list[i]);
        }
    }
}

int main()
{
    int qwq[10]={1,2,3,4,5};
    Pern(qwq,0,5);
}

 

7.二分搜索

 

#include<bits/stdc++.h>
using namespace std;

//二分查找 arr是数组名，len是长度，key是要查找的数
//查到则返回元素下标 查不到返回-1
int binarySearch(int arr[], int len, int key)
{
    int left = 0;
    int right = len - 1;
    int mid;

    while (left <= right) {
        mid = (left + right) / 2;
        if (key < arr[mid]) {//key在左边
            right = mid - 1;
        } else if (arr[mid] < key) {//key在右边
            left = mid + 1;
        } else {
            return mid;
        }
    }
    return -1;
}

int main()
{
    int qwq[5]={1,5,22,132,133};
    //在长度为5的qwq数组中查找22
    cout<<binarySearch(qwq,5,22);

}

 

数据结构

 

并查集

8.并查集

int father[10005];
void inti()
{
    //初始化 father[i] = i;
}
//递归写法
int findd(int x)
{
    if(father[x] == x) return x;
    else
    {
        father[x] = findd(father[x]);
        return father[x];
    }
}
//迭代写法
/*
int findd(int x) {   //  迭代找根节点
    int root = x; // 根节点
    while (root != father[root]) { // 寻找根节点
        root = father[root];
    }
    while (x != root) {
        int tmp = father[x];
        father[x] = root; // 根节点赋值
        x = tmp;
    }
    return root;
}
*/
void join(int x,int y)
{
    int fx = findd(x);
    int fy = findd(y);
    if(fx != fy) father[fy] = fx;
}

 

图论

 

MST

最小生成树

Kruskal

9.克鲁斯卡尔算法

/*
    |Kruskal算法|
    |适用于 稀疏图 求最小生成树|
    |16/11/05ztx thanks to wangqiqi|
*/

/*
    第一步：点、边、加入vector，把所有边按从小到大排序
    第二步：并查集部分 + 下面的code
*/

void Kruskal() {    
    ans = 0;    
    for (int i = 0; i<len; i++) {    
        if (Find(edge[i].a) != Find(edge[i].b)) {    
            Union(edge[i].a, edge[i].b);    
            ans += edge[i].len;    
        }    
    }    
}    

 

•  

 

Prim

10.普里姆算法

/*
    |Prim算法|
    |适用于 稠密图 求最小生成树|
    |堆优化版，时间复杂度：O(elgn)|
    |16/11/05ztx, thanks to chaixiaojun|
*/

struct node {  
    int v, len;  
    node(int v = 0, int len = 0) :v(v), len(len) {}  
    bool operator < (const node &a)const {  // 加入队列的元素自动按距离从小到大排序  
        return len> a.len;  
    }  
};

vector<node> G[maxn];
int vis[maxn];
int dis[maxn];

void init() {  
    for (int i = 0; i<maxn; i++) {  
        G[i].clear();  
        dis[i] = INF;  
        vis[i] = false;  
    }  
}  
int Prim(int s) {  
    priority_queue<node>Q; // 定义优先队列  
    int ans = 0;  
    Q.push(node(s,0));  // 起点加入队列  
    while (!Q.empty()) {   
        node now = Q.top(); Q.pop();  // 取出距离最小的点  
        int v = now.v;  
        if (vis[v]) continue;  // 同一个节点，可能会推入2次或2次以上队列，这样第一个被标记后，剩下的需要直接跳过。  
        vis[v] = true;  // 标记一下  
        ans += now.len;  
        for (int i = 0; i<G[v].size(); i++) {  // 开始更新  
            int v2 = G[v][i].v;  
            int len = G[v][i].len;  
            if (!vis[v2] && dis[v2] > len) {   
                dis[v2] = len;  
                Q.push(node(v2, dis[v2]));  // 更新的点加入队列并排序  
            }  
        }  
    }  
    return ans; 
}  

#include<iostream>
#include<string>
#include<vector>
using  namespace std;

//首先是使用邻接矩阵完成Prim算法
struct Graph {
    int vexnum;  //顶点个数
    int edge;   //边的条数
    int ** arc; //邻接矩阵
    string *information; //记录每个顶点名称
};

//创建图
void createGraph(Graph & g) {
    cout << "请输入顶点数：输入边的条数" << endl;
    cin >> g.vexnum;
    cin >> g.edge;  //输入边的条数

    g.information = new string[g.vexnum];
    g.arc = new int*[g.vexnum];
    int i = 0;

    //开辟空间的同时，进行名称的初始化
    for (i = 0; i < g.vexnum; i++) {
        g.arc[i] = new int[g.vexnum];
        g.information[i]="v"+ std::to_string(i+1);//对每个顶点进行命名
        for (int k = 0; k < g.vexnum; k++) {
            g.arc[i][k] = INT_MAX;          //初始化我们的邻接矩阵
        }
    }

    cout << "请输入每条边之间的顶点编号(顶点编号从1开始),以及该边的权重：" << endl;
    for (i = 0; i < g.edge; i++) {
        int start;
        int end;
        cin >> start;   //输入每条边的起点
        cin >> end;     //输入每条边的终点
        int weight;
        cin >> weight;
        g.arc[start-1][end-1]=weight;//无向图的边是相反的
        g.arc[end-1][start-1] = weight;
    }
}

//打印图
void print(Graph g) {
    int i;
    for (i = 0; i < g.vexnum; i++) {
        //cout << g.information[i] << " ";
        for (int j = 0; j < g.vexnum; j++) {
            if (g.arc[i][j] == INT_MAX)
                cout << "∞" << " ";
            else
            cout << g.arc[i][j] << " ";
        }
        cout << endl;
    }
}

//作为记录边的信息，这些边都是达到end的所有边中，权重最小的那个
struct Assis_array {
    int start; //边的终点
    int end;  //边的起点
    int weight;  //边的权重
};
//进行prim算法实现，使用的邻接矩阵的方法实现。
void Prim(Graph g,int begin) {

    //close_edge这个数组记录到达某个顶点的各个边中的权重最大的那个边
    Assis_array *close_edge=new Assis_array[g.vexnum];

    int j;

    //进行close_edge的初始化，更加开始起点进行初始化
    for (j = 0; j < g.vexnum; j++) {
        if (j != begin - 1) {
            close_edge[j].start = begin-1;
            close_edge[j].end = j;
            close_edge[j].weight = g.arc[begin - 1][j];
        }
    }
    //把起点的close_edge中的值设置为-1，代表已经加入到集合U了
    close_edge[begin - 1].weight = -1;
    //访问剩下的顶点，并加入依次加入到集合U
    for (j = 1; j < g.vexnum; j++) {

        int min = INT_MAX;
        int k;
        int index;
        //寻找数组close_edge中权重最小的那个边
        for (k = 0; k < g.vexnum; k++) {
            if (close_edge[k].weight != -1) {  
                if (close_edge[k].weight < min) {
                    min = close_edge[k].weight;
                    index = k;
                }
            }
        }
        //将权重最小的那条边的终点也加入到集合U
        close_edge[index].weight = -1;
        //输出对应的边的信息
        cout << g.information[close_edge[index].start] 
            << "-----" 
            << g.information[close_edge[index].end]
            << "="
            <<g.arc[close_edge[index].start][close_edge[index].end]
            <<endl;

        //更新我们的close_edge数组。
        for (k = 0; k < g.vexnum; k++) {
            if (g.arc[close_edge[index].end][k] <close_edge[k].weight) {
                close_edge[k].weight = g.arc[close_edge[index].end][k];
                close_edge[k].start = close_edge[index].end;
                close_edge[k].end = k;
            }
        }
    }
}



int main()
{
    Graph g;
    createGraph(g);//基本都是无向网图，所以我们只实现了无向网图
    print(g);
    Prim(g, 1);
    system("pause");
    return 0;
}

   

 

Bellman-Ford

 

单源最短路

Dijkstra

11.迪杰斯特拉算法

#include<bits/stdc++.h>
using namespace std;


int qwq[1005][1005];
int n,m,a,b,c;
int dis[1005];
int book[1005];
int maxn=65355;

int main()
{
    while(scanf("%d%d",&n,&m),n,m){


        memset(qwq,0,sizeof(qwq));
        memset(book,0,sizeof(book));
        memset(dis,0,sizeof(dis));
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            if(i==j) qwq[i][j]=0;
            else qwq[i][j]=maxn;
        }


           for(int i=1;i<=m;i++){

            scanf("%d%d%d",&a,&b,&c);
            qwq[a][b]=qwq[b][a]=c;
        }


  for(int i=1;i<=n;i++){

            dis[i]=qwq[1][i];
        }

        book[1]=1;
		int min,u;





        for(int i=1;i<=n;i++)

        {
             min=maxn;
             for(int j=1;j<=n;j++){


                if(book[j]==0 && min>dis[j])
                {
                    min=dis[j];
                    u=j;
                }

             }
                book[u]=1;

               for(int j=1;j<=n;j++)
               {

                    if(book[j]==0 && dis[j]>dis[u]+qwq[u][j])
                    dis[j]=dis[u]+qwq[u][j];
               }
        }


        cout<<dis[n]<<endl;
    }

}

 

SPFA

12.最短路径快速算法（Shortest Path Faster Algorithm）

#include<bits/stdc++.h>
using namespace std;

const int MAXN = 110;
const int MAXM = 10100;
const int INF = 0x3f3f3f3f;

struct Edge
{
    int from, to, cap, next;
};

Edge edge[MAXM];
int head[MAXN];
int path[MAXN];
int inqueue[MAXN];
int dist[MAXN];
int viscnt[MAXN];
int cnt;

void addedge( int from, int to, int cap )
{
    edge[cnt].from = from;
    edge[cnt].to = to;
    edge[cnt].cap = cap;
    edge[cnt].next = head[from];
    head[from] = cnt++;
}

int relax(int u,int v,int c)
{
    if (dist[u] + c < dist[v])
    {
        dist[v] = dist[u] + c;
        return 1;
    }
    return 0;
}

bool SPFA( int src, int n )
{
    deque<int> dq;
    memset( viscnt, 0, sizeof viscnt );
    memset( inqueue, 0, sizeof inqueue );
    memset( dist, INF, sizeof dist );
    memset( path, -1, sizeof path );
    inqueue[src] = 1;
    viscnt[src]++;
    dist[src] = 0;

    dq.push_back( src );
    while (!dq.empty())
    {
        int u = dq.front();
        dq.pop_front();
        inqueue[u] = 0;

        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dist[u] < INF&&relax( u, v, edge[i].cap ))
            {
                path[v] = u;
                if(!inqueue[v])
                {
                    inqueue[v] = 1;
                    viscnt[v]++;
                    if(viscnt[v] == n)  return false;
                    if(!dq.empty() && dist[v] <= dist[dq.front()])
                        dq.push_front( v );
                    else
                        dq.push_back( v );
                }
            }
        }
    }
    return true;
}

int main()
{
    int n,m;
    while(cin >> n >> m &&n&&m)
    {
        memset( head, -1, sizeof head );
        cnt = 0;
        for(int i = 1; i <= m; i++)
        {
            int a, b, c;
            cin >> a >> b >> c;
            addedge( a, b, c );//在a->b添加一条负载为c的边
            addedge( b, a, c );
        }
        SPFA( 1, n );
        cout << dist[n] << endl;
    }
    return 0;
}

 

Floyd-Warshall

13.弗洛伊德算法

/*
    |Floyd算法|
    |任意点对最短路算法|
    |求图中任意两点的最短距离的算法|
*/

for (int k = 0; k < n; k++) {  
    for (int i = 0; i < n; i++) {  
        for (int j = 0; j < n; j++) {  
            dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);  
        }  
    }
}

 

二分图

 

14.染色法

/*
    |交叉染色法判断二分图|
    |16/11/05ztx|
*/

int bipartite(int s) {  
    int u, v;  
    queue<int>Q;  
    color[s] = 1;  
    Q.push(s);  
    while (!Q.empty()) {  
        u = Q.front();  
        Q.pop();  
        for (int i = 0; i < G[u].size(); i++) {  
            v = G[u][i];  
            if (color[v] == 0) {  
                color[v] = -color[u];  
                Q.push(v);  
            }  
            else if (color[v] == color[u])  
                return 0;  
        }  
    }  
    return 1;  
}  

 

15..匈牙利算法

 

/*
    |求解最大匹配问题|
    |递归实现|
    |16/11/05ztx|
*/

vector<int>G[maxn];  
bool inpath[maxn];  //  标记  
int match[maxn];    //  记录匹配对象  
void init()  
{  
    memset(match, -1, sizeof(match));  
    for (int i = 0; i < maxn; ++i) {  
        G[i].clear();  
    }  
}  
bool findpath(int k) {  
    for (int i = 0; i < G[k].size(); ++i) {  
        int v = G[k][i];  
        if (!inpath[v]) {  
            inpath[v] = true;  
            if (match[v] == -1 || findpath(match[v])) { // 递归  
                match[v] = k; // 即匹配对象是“k妹子”的  
                return true;  
            }  
        }  
    }  
    return false;  
}  

void hungary() {  
    int cnt = 0;  
    for (int i = 1; i <= m; i++) {  // m为需要匹配的“妹子”数  
        memset(inpath, false, sizeof(inpath)); // 每次都要初始化  
        if (findpath(i)) cnt++;  
    }  
    cout << cnt << endl;  
}  

/*
    |求解最大匹配问题|
    |dfs实现|
    |16/11/05ztx|
*/

int v1, v2;  
bool Map[501][501];  
bool visit[501];  
int link[501];  
int result;  

bool dfs(int x)  {  
    for (int y = 1; y <= v2; ++y)  {  
        if (Map[x][y] && !visit[y])  {  
            visit[y] = true;  
            if (link[y] == 0 || dfs(link[y]))  {  
                link[y] = x;  
                return true;  
            } } }  
    return false;  
}  


void Search()  {  
    for (int x = 1; x <= v1; x++)  {  
        memset(visit,false,sizeof(visit));  
        if (dfs(x))  
            result++;  
    }
}

 

 

 

动态规划

背包

16.17.18背包问题

/*
    |01背包|
    |完全背包|
    |多重背包|
    |16/11/05ztx|
*/

//  01背包：  

void bag01(int cost,int weight)  {  
    for(i = v; i >= cost; --i)  
    dp[i] = max(dp[i], dp[i-cost]+weight);  
}  

//  完全背包：  

void complete(int cost, int weight)  {  
    for(i = cost ; i <= v; ++i)  
    dp[i] = max(dp[i], dp[i - cost] + weight);  
}  

//  多重背包：  

void multiply(int cost, int weight, int amount)  {  
    if(cost * amount >= v)  
        complete(cost, weight);  
    else{  
        k = 1;  
        while (k < amount){  
            bag01(k * cost, k * weight);  
            amount -= k;  
            k += k;  
        }  
        bag01(cost * amount, weight * amount);  
    }  
}  


// other

int dp[1000000];
int c[55], m[110];
int sum;

void CompletePack(int c) {
    for (int v = c; v <= sum / 2; ++v){
        dp[v] = max(dp[v], dp[v - c] + c);
    }
}

void ZeroOnePack(int c) {
    for (int v = sum / 2; v >= c; --v) {
        dp[v] = max(dp[v], dp[v - c] + c);
    }
}

void multiplePack(int c, int m） {
    if (m * c > sum / 2)
        CompletePack(c);
    else{
        int k = 1;
        while (k < m){
            ZeroOnePack(k * c);
            m -= k;
            k <<= 1;
        }
        if (m != 0){
            ZeroOnePack(m * c);
        }
    }
}

 

LIS

 

19.最长上升子序列

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;

const int N = 131072;

int n = 7, a[N] = {0,0,1,1,0,0,2};

template<class Cmp>
int LIS (Cmp cmp)
{
    static int m, end[N];
    m = 0;
    for (int i=0;i<n;i++)
    {
        int pos = lower_bound(end, end+m, a[i], cmp)-end;
        end[pos] = a[i], m += pos==m;
    }
    return m;
}
bool greater1(int value)
{
    return value >=1;
}

int main ()
{
    cin>>n;
    for(int i=0;i<n;i++) scanf("%d",&a[i]);
    cout << LIS(less<int>()) << endl;         //严格上升
    cout << LIS(less_equal<int>()) << endl;   //非严格上升
    cout << LIS(greater<int>()) << endl;      //严格下降
    cout << LIS(greater_equal<int>()) << endl;//非严格下降
    cout << count_if(a,a+7,greater1) << endl; //计数
    //第二个参数为末尾元素的下一个位置
    return 0;
}
//OUT：
//3 5 2 4 3

 

LCS

 

20.最长公共子序列

#include<bits/stdc++.h>
using namespace std;
int dp[1005][1005];
char s1[1005],s2[1005];
int main()
{
    int len1,len2,i,j;
    cin>>s1>>s2;
    len1=strlen(s1);
    len2=strlen(s2);
    memset(dp,0,sizeof(dp));
    for(i=1;i<=len1;i++)
        for(j=1;j<=len2;j++)
        {
            if(s1[i-1]==s2[j-1])
                dp[i][j]=dp[i-1][j-1]+1;
            else if(dp[i-1][j]>=dp[i][j-1])
                dp[i][j]=dp[i-1][j];
            else
                dp[i][j]=dp[i][j-1];
        }
    cout<< dp[len1][len2];
}

 

计算几何

 

21.向量基本用法

/*
    |16/11/06ztx|
*/

struct node {  
    double x; // 横坐标  
    double y; // 纵坐标  
};  

typedef node Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }  
Vector operator - (Point A, Point B) { return Vector(A.x - B.y, A.y - B.y); }  
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }  
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y*p); }  

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } // 向量点乘  
double Length(Vector A) { return sqrt(Dot(A, A)); }  // 向量模长  
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }  // 向量之间夹角  

double Cross(Vector A, Vector B) { // 叉积计算 公式  
    return A.x*B.y - A.y*B.x;  
}  

Vector Rotate(Vector A, double rad) // 向量旋转 公式  {  
    return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));  
}  

Point getLineIntersection(Point P, Vector v, Point Q, Vector w) { // 两直线交点t1 t2计算公式   
    Vector u = P - Q;   
    double t = Cross(w, u) / Cross(v, w);  // 求得是横坐标  
    return P + v*t;  // 返回一个点  
}  

 

22.求多边形面积

 

/*
    |16/11/06ztx|
*/

node G[maxn];  
int n;  

double Cross(node a, node b) { // 叉积计算  
    return a.x*b.y - a.y*b.x;  
}  


int main()  
{  
    while (scanf("%d", &n) != EOF && n) {  
        for (int i = 0; i < n; i++)   
            scanf("%lf %lf", &G[i].x, &G[i].y);  
        double sum = 0;  
        G[n].x = G[0].x;  
        G[n].y = G[0].y;  
        for (int i = 0; i < n; i++) {   
                sum += Cross(G[i], G[i + 1]);  
        }  
        // 或者  
            //for (int i = 0; i < n; i++) {  
                //sum += fun(G[i], G[（i + 1）% n]);  
            //}  
        sum = sum / 2.0;  
        printf("%.1f\n", sum);  
    }  
    system("pause");  
    return 0;  
}

 

23..判断线段相交

 

/*
    |16/11/06ztx|
*/

node P[35][105];     

double Cross_Prouct(node A,node B,node C) {     //  计算BA叉乘CA     
    return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);      
}      
bool Intersect(node A,node B,node C,node D)  {  //  通过叉乘判断线段是否相交；           
    if(min(A.x,B.x)<=max(C.x,D.x)&&         //  快速排斥实验；      
       min(C.x,D.x)<=max(A.x,B.x)&&      
       min(A.y,B.y)<=max(C.y,D.y)&&      
       min(C.y,D.y)<=max(A.y,B.y)&&      
       Cross_Prouct(A,B,C)*Cross_Prouct(A,B,D)<0&&      //  跨立实验；      
       Cross_Prouct(C,D,A)*Cross_Prouct(C,D,B)<0)       //  叉乘异号表示在两侧；      
       return true;      
    else return false;      
}    

24.求三角形外心
 

/*
    |16/11/06ztx|
*/

Point circumcenter(const Point &a, const Point &b, const Point &c) { //返回三角形的外心        
    Point ret;  
    double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1) / 2;  
    double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2) / 2;  
    double d = a1*b2 - a2*b1;  
    ret.x = a.x + (c1*b2 - c2*b1) / d;  
    ret.y = a.y + (a1*c2 - a2*c1) / d;  
    return ret;  
}  

 

24.极角排序

 

/*
    |16/11/06ztx|
*/

double cross(point p1, point p2, point q1, point q2) {  // 叉积计算   
    return (q2.y - q1.y)*(p2.x - p1.x) - (q2.x - q1.x)*(p2.y - p1.y);  
}  
bool cmp(point a, point b)  {  
    point o;  
    o.x = o.y = 0;  
    return cross(o, b, o, a) < 0; // 叉积判断  
}  
sort(convex + 1, convex + cnt, cmp); // 按角排序, 从小到大 

 

字符串

 

kmp

25.克努特-莫里斯-普拉特操作

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100

void cal_next( char * str, int * next, int len )
{
    int i, j;

    next[0] = -1;
    for( i = 1; i < len; i++ )
    {
        j = next[ i - 1 ];
        while( str[ j + 1 ] != str[ i ] && ( j >= 0 ) )
        {
            j = next[ j ];
        }
        if( str[ i ] == str[ j + 1 ] )
        {
            next[ i ] = j + 1;
        }
        else
        {
            next[ i ] = -1;
        }
    }
}

int KMP( char * str, int slen, char * ptr, int plen, int * next )
{
    int s_i = 0, p_i = 0;

    while( s_i < slen && p_i < plen )
    {
        if( str[ s_i ] == ptr[ p_i ] )
        {
            s_i++;
            p_i++;
        }
        else
        {
            if( p_i == 0 )
            {
                s_i++;
            }
            else
            {
                p_i = next[ p_i - 1 ] + 1;
            }
        }
    }
    return ( p_i == plen ) ? ( s_i - plen ) : -1;
}

int main()
{
    char str[ N ] = {0};
    char ptr[ N ] = {0};
    int slen, plen;
    int next[ N ];

    while( scanf( "%s%s", str, ptr ) )
    {
        slen = strlen( str );
        plen = strlen( ptr );
        cal_next( ptr, next, plen );
        printf( "%d\n", KMP( str, slen, ptr, plen, next ) );
    }
    return 0;
}

 

26.kmp扩展

 

/*
    |16/11/06ztx|
*/

#include<iostream>    
#include<cstring>    

using namespace std;

const int MM=100005;    

int next[MM],extand[MM];    
char S[MM],T[MM];    

void GetNext(const char *T) {    
    int len = strlen(T),a = 0;    
    next[0] = len;    
    while(a < len - 1 && T[a] == T[a + 1]) a++;    
    next[1] = a;    
    a = 1;    
    for(int k = 2; k < len; k ++) {    
        int p = a + next[a] - 1,L = next[k - a];    
        if( (k - 1) + L >= p) {    
            int j = (p - k + 1) > 0 ? (p - k + 1) : 0;    
            while(k + j < len && T[k + j] == T[j]) j++;    
            next[k] = j;    
            a = k;    
        }else next[k] = L;    
    }    
}    
void GetExtand(const char *S,const char *T) {    
    GetNext(T);    
    int slen = strlen(S),tlen = strlen(T),a = 0;    
    int MinLen = slen < tlen ? slen : tlen;    
    while(a < MinLen && S[a] == T[a]) a++;    
    extand[0] = a;     
    a = 0;    
    for(int k = 1; k < slen; k ++) {    
        int p = a + extand[a] - 1, L = next[k - a];    
        if( (k - 1) + L >= p) {    
            int j = (p - k + 1) > 0 ? (p - k + 1) : 0;    
            while(k + j < slen && j < tlen && S[k + j] == T[j]) j ++;    
            extand[k] = j;    
            a = k;    
        } else    
            extand[k] = L;    
    }    
}    
void show(const int *s,int len){    
    for(int i = 0; i < len; i ++)    
            cout << s[i] << ' ';    
    cout << endl;    
}    

int main() {    
    while(cin >> S >> T) {    
        GetExtand(S,T);    
        show(next,strlen(T));    
        show(extand,strlen(S));    
    }    
    return 0;    
}   

 

字典树

 

27.字典树

/*
    |16/11/06ztx|
*/

struct Trie{  
    int cnt;  
    Trie *next[maxn];  
    Trie(){  
        cnt = 0;  
        memset(next,0,sizeof(next));  
    }  
};  

Trie *root;  

void Insert(char *word)  {  
    Trie *tem = root;  
    while(*word != '\0')  {  
        int x = *word - 'a';  
        if(tem->next[x] == NULL)  
            tem->next[x] = new Trie;  
        tem = tem->next[x];  
        tem->cnt++;  
        word++;  
    }  
}  

int Search(char *word)  {  
    Trie *tem = root;  
    for(int i=0;word[i]!='\0';i++)  {  
        int x = word[i]-'a';  
        if(tem->next[x] == NULL)  
            return 0;  
        tem = tem->next[x];  
    }  
    return tem->cnt;  
}  

void Delete(char *word,int t) {  
    Trie *tem = root;  
    for(int i=0;word[i]!='\0';i++)  {  
        int x = word[i]-'a';  
        tem = tem->next[x];  
        (tem->cnt)-=t;  
    }  
    for(int i=0;i<maxn;i++)  
        tem->next[i] = NULL;  
}  

int main() {  
    int n;  
    char str1[50];  
    char str2[50];  
    while(scanf("%d",&n)!=EOF)  {  
        root = new Trie;  
        while(n--)  {  
            scanf("%s %s",str1,str2);  
            if(str1[0]=='i') {
                Insert(str2); 
            }else if(str1[0] == 's')  {  
                if(Search(str2))  
                    printf("Yes\n");  
                else  
                    printf("No\n");  
            }else  {  
                int t = Search(str2);  
                if(t)  
                    Delete(str2,t);  
            } } }  
    return 0;  
}  

 

28.AC自动机

 

/*
    |16/11/06ztx|
*/

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<string>  

using namespace std;  

#define N 1000010  

char str[N], keyword[N];  
int head, tail;  

struct node {  
    node *fail;  
    node *next[26];  
    int count;  
    node() { //init  
        fail = NULL;// 默认为空  
        count = 0;  
        for(int i = 0; i < 26; ++i)  
            next[i] = NULL;  
    }  
}*q[N];  

node *root;  

void insert(char *str)  { // 建立Trie  
    int temp, len;  
    node *p = root;  
    len = strlen(str);  
    for(int i = 0; i < len; ++i)  {  
        temp = str[i] - 'a';  
        if(p->next[temp] == NULL)  
            p->next[temp] = new node();  
        p = p->next[temp];  
    }  
    p->count++;  
}  

void build_ac() { // 初始化fail指针，BFS 数组模拟队列：   
    q[tail++] = root;  
    while(head != tail)  {  
        node *p = q[head++]; // 弹出队头  
        node *temp = NULL;  
        for(int i = 0; i < 26; ++i)  {  
            if(p->next[i] != NULL)  {  
                if(p == root) { // 第一个元素fail必指向根  
                    p->next[i]->fail = root;
                }else {  
                    temp = p->fail; // 失败指针  
                    while(temp != NULL) { // 2种情况结束：匹配为空or找到匹配 
                        if(temp->next[i] != NULL) { // 找到匹配  
                            p->next[i]->fail = temp->next[i];  
                            break;  
                        }  
                        temp = temp->fail;  
                    }  
                    if(temp == NULL) // 为空则从头匹配  
                        p->next[i]->fail = root;  
                }  
                q[tail++] = p->next[i]; // 入队  
            } } }  
}  

int query() // 扫描  
{  
    int index, len, result;  
    node *p = root; // Tire入口  
    result = 0;  
    len = strlen(str);  
    for(int i = 0; i < len; ++i)  
    {  
        index = str[i] - 'a';  
        while(p->next[index] == NULL && p != root) // 跳转失败指针  
            p = p->fail;  
        p = p->next[index];  
        if(p == NULL)  
            p = root;  
        node *temp = p; // p不动，temp计算后缀串  
        while(temp != root && temp->count != -1)   {  
            result += temp->count;  
            temp->count = -1;  
            temp = temp->fail;  
        }  
    }  
    return result;  
}  

int main() {  
    int num;  
    head= tail = 0;  
    root = new node();  
    scanf("%d", &num);  
    getchar();  
    for(int i = 0; i < num; ++i) {  
       scanf("%s",keyword);  
        insert(keyword);  
    }  
    build_ac();  
    scanf("%s", str);  
    if(query())  
        printf("YES\n");  
    else  
        printf("NO\n");  
    return 0;  
}  

/*
    假设有N个模式串，平均长度为L；文章长度为M。 建立Trie树：O(N*L) 建立fail指针：O(N*L) 模式匹配：O(M*L) 所以，总时间复杂度为:O( (N+M)*L )。
*/

 

线段树

 

29.线段树 

1）点更新

/*
    |16/12/07ztx|
*/

struct node
{
    int left, right;
    int max, sum;
};

node tree[maxn << 2];
int a[maxn];
int n;
int k = 1;
int p, q;
string str;

void build(int m, int l, int r)//m 是 树的标号
{
    tree[m].left = l;
    tree[m].right = r;
    if (l == r){
        tree[m].max = a[l];
        tree[m].sum = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(m << 1, l, mid);
    build(m << 1 | 1, mid + 1, r);
    tree[m].max = max(tree[m << 1].max, tree[m << 1 | 1].max);
    tree[m].sum = tree[m << 1].sum + tree[m << 1 | 1].sum;
}

void update(int m, int a, int val)//a 是 节点位置， val 是 更新的值（加减的值）
{
    if (tree[m].left == a && tree[m].right == a){
        tree[m].max += val;
        tree[m].sum += val;
        return;
    }
    int mid = (tree[m].left + tree[m].right) >> 1;
    if (a <= mid){
        update(m << 1, a, val);
    }
    else{
        update(m << 1 | 1, a, val);
    }
    tree[m].max = max(tree[m << 1].max, tree[m << 1 | 1].max);
    tree[m].sum = tree[m << 1].sum + tree[m << 1 | 1].sum;
}

int querySum(int m, int l, int r)
{
    if (l == tree[m].left && r == tree[m].right){
        return tree[m].sum;
    }
    int mid = (tree[m].left + tree[m].right) >> 1;
    if (r <= mid){
        return querySum(m << 1, l, r);
    }
    else if (l > mid){
        return querySum(m << 1 | 1, l, r);
    }
    return querySum(m << 1, l, mid) + querySum(m << 1 | 1, mid + 1, r);
}

int queryMax(int m, int l, int r)
{
    if (l == tree[m].left && r == tree[m].right){
        return tree[m].max;
    }
    int mid = (tree[m].left + tree[m].right) >> 1;
    if (r <= mid){
        return queryMax(m << 1, l, r);
    }
    else if (l > mid){
        return queryMax(m << 1 | 1, l, r);
    }
    return max(queryMax(m << 1, l, mid), queryMax(m << 1 | 1, mid + 1, r));
} 

build(1,1,n);  
update(1,a,b);  
query(1,a,b);  

 

2)区间更新

 

/*
    |16/11/06ztx|
*/

typedef long long ll;  
const int maxn = 100010;  

int t,n,q;  
ll anssum;  

struct node{  
    ll l,r;  
    ll addv,sum;  
}tree[maxn<<2];  

void maintain(int id) {  
    if(tree[id].l >= tree[id].r)  
        return ;  
    tree[id].sum = tree[id<<1].sum + tree[id<<1|1].sum;  
}  

void pushdown(int id) {  
    if(tree[id].l >= tree[id].r)  
        return ;  
    if(tree[id].addv){  
        int tmp = tree[id].addv;  
        tree[id<<1].addv += tmp;  
        tree[id<<1|1].addv += tmp;  
        tree[id<<1].sum += (tree[id<<1].r - tree[id<<1].l + 1)*tmp;  
        tree[id<<1|1].sum += (tree[id<<1|1].r - tree[id<<1|1].l + 1)*tmp;  
        tree[id].addv = 0;  
    }  
}  

void build(int id,ll l,ll r) {  
    tree[id].l = l;  
    tree[id].r = r;  
    tree[id].addv = 0;  
    tree[id].sum = 0;  
    if(l==r)  {  
        tree[id].sum = 0;  
        return ;  
    }  
    ll mid = (l+r)>>1;  
    build(id<<1,l,mid);  
    build(id<<1|1,mid+1,r);  
    maintain(id);  
}  

void updateAdd(int id,ll l,ll r,ll val) {  
    if(tree[id].l >= l && tree[id].r <= r)  
    {  
        tree[id].addv += val;  
        tree[id].sum += (tree[id].r - tree[id].l+1)*val;  
        return ;  
    }  
    pushdown(id);  
    ll mid = (tree[id].l+tree[id].r)>>1;  
    if(l <= mid)  
        updateAdd(id<<1,l,r,val);  
    if(mid < r)  
        updateAdd(id<<1|1,l,r,val);  
    maintain(id);  
}  

void query(int id,ll l,ll r) {  
    if(tree[id].l >= l && tree[id].r <= r){  
        anssum += tree[id].sum;  
        return ;  
    }  
    pushdown(id);  
    ll mid = (tree[id].l + tree[id].r)>>1;  
    if(l <= mid)  
        query(id<<1,l,r);  
    if(mid < r)  
        query(id<<1|1,l,r);  
    maintain(id);  
}  

int main() {  
    scanf("%d",&t);  
    int kase = 0 ;  
    while(t--){  
        scanf("%d %d",&n,&q);  
        build(1,1,n);  
        int id;  
        ll x,y;  
        ll val;  
        printf("Case %d:\n",++kase);  
        while(q--){  
            scanf("%d",&id);  
            if(id==0){  
                scanf("%lld %lld %lld",&x,&y,&val);  
                updateAdd(1,x+1,y+1,val);  
            }  
            else{  
                scanf("%lld %lld",&x,&y);  
                anssum = 0;  
                query(1,x+1,y+1);  
                printf("%lld\n",anssum);  
            } } }  
    return 0;  
}  

 

30.树状数组

 

/*
    |16/11/06ztx|
*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>

using namespace std;

typedef long long ll;

const int maxn = 50005;

int a[maxn];
int n;

int lowbit(const int t) {
    return t & (-t);
}

void insert(int t, int d) {
    while (t <= n){
        a[t] += d;
        t = t + lowbit(t);
    }
}

ll getSum(int t) {
    ll sum = 0;
    while (t > 0){
        sum += a[t];
        t = t - lowbit(t);
    }
    return sum;
}

int main() {
    int t, k, d;
    scanf("%d", &t);
    k= 1;
    while (t--){
        memset(a, 0, sizeof(a));
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &d);
            insert(i, d);
        }
        string str;
        printf("Case %d:\n", k++);
        while (cin >> str) {
            if (str == "End")   break;
            int x, y;
            scanf("%d %d", &x, &y);
            if (str == "Query")
                printf("%lld\n", getSum(y) - getSum(x - 1));
            else if (str == "Add")
                insert(x, y);
            else if (str == "Sub")
                insert(x, -y);
        }
    }
    return 0;
}

 

其他

 

31.中国剩余定理（孙子定理）

/*
    |16/11/06ztx|
*/

int CRT(int a[],int m[],int n)  {    
    int M = 1;    
    int ans = 0;    
    for(int i=1; i<=n; i++)    
        M *= m[i];    
    for(int i=1; i<=n; i++)  {    
        int x, y;    
        int Mi = M / m[i];    
        extend_Euclid(Mi, m[i], x, y);    
        ans = (ans + Mi * x * a[i]) % M;    
    }    
    if(ans < 0) ans += M;    
    return ans;    
}  

void extend_Euclid(int a, int b, int &x, int &y)  {  
    if(b == 0) {  
        x = 1;  
        y = 0;  
        return;  
    }  
    extend_Euclid(b, a % b, x, y);  
    int tmp = x;  
    x = y;  
    y = tmp - (a / b) * y;  
}  

32.判断字符串是否包含某个子串

#include<bits/stdc++.h>  
using namespace std;  
  
int main()  
{     
    //定义两个相同部分为hello的字符串  
    string str1="asdasdadhelloz,zxc";  
    string str2="hello";  
     
   //应该是一个类似迭代器的东西…还没有研究透  
    string::size_type id = str1.find(str2);  
      
    if(id != string::npos )  
    {  
        cout<<"字符串中含有"<<str2<<endl;  
    }  
      
    else   
    {  
        cout<<"字符串中不含有"<<str2<<endl;  
    }  
}  

33.模拟二(n)进制相加

#include<bits/stdc++.h>
using namespace std;

char a[105],b[105];
int ans[105];


int main()
{
    cin>>a>>b;
    int len = strlen(a);
    for(int i=len-1;i>0;i--)
    {
        ans[i] +=a[i]-48 + b[i] -48;
        //开始模拟进位
        ans[i-1] += ans[i] /2;
        ans[i] = ans[i]%2;
    }
    ans[0] += a[0] -48+b[0] -48;

    if(ans[0] > 1) cout<<ans[0]/2<<ans[0] %2;

    else cout<<ans[0];

    for(int i=1;i<len;i++)
    {
        cout<<ans[i];
    }
}

34.

Wannafly模拟赛 矩阵 [矩阵hash+二分]

 

题目描述

给出一个n * m的矩阵。让你从中发现一个最大的正方形。使得这样子的正方形在矩阵中出现了至少两次。输出最大正方形的边长。

输入描述:

第一行两个整数n, m代表矩阵的长和宽；
接下来n行，每行m个字符（小写字母），表示矩阵；

输出描述:

输出一个整数表示满足条件的最大正方形的边长。

示例1

输入

5 10
ljkfghdfas
isdfjksiye
pgljkijlgp
eyisdafdsi
lnpglkfkjl

输出

3

备注:

对于30%的数据，n,m≤100；
对于100%的数据，n,m≤500；

 

 

 

思路：

 

二分答案，然后Hash一下就行。

 

Ac代码：

 

#include<stdio.h>  
#include<string.h>  
#include<map>  
#include<algorithm>  
#include<iostream>  
using namespace std;  
char a[505][505];  
unsigned long long int h[505][505];  
unsigned long long int X[505*505];  
int n,m;  
const int maxn=350000;  
const int hashh=1000007;  
   
struct hashmap  
{  
    unsigned long long int a[maxn];  
    int head[hashh];  
    int next[maxn];  
    int size;  
    void init()  
    {  
        memset(head,-1,sizeof(head));  
        size=0;  
    }  
    bool find(unsigned long long int val)  
    {  
        int tmp=(val%hashh+hashh)%hashh;  
        for(int i=head[tmp];i!=-1;i=next[i])  
        {  
            if(val==a[i])return true;  
        }  
        return false;  
    }  
    void add(unsigned long long int val)  
    {  
        int tmp=(val%hashh+hashh)%hashh;  
        if(find(val))return ;  
        a[size]=val;  
        next[size]=head[tmp];//令next指向-1、  
        head[tmp]=size++;  
    }  
}H;  
int Slove(int mid)  
{  
    H.init();  
    map<unsigned long long int,int>s;  
    for(int i=1;i<=n;i++)  
    {  
        for(int j=1;j<=m;j++)  
        {  
            int x=i,y=j;  
            int xx=i+mid-1,yy=j+mid-1;  
            if(xx>=1&&xx<=n&&yy>=1&&yy<=m)  
            {  
                unsigned long long int val=h[xx][yy]+h[x-1][y-1]-h[xx][y-1]-h[x-1][yy];  
                val*=X[(n-i)*m+m-j];  
                if(H.find(val)==true)return 1;  
                else H.add(val);  
            }  
        }  
    }  
    return 0;  
}  
int main()  
{  
    while(~scanf("%d%d",&n,&m))  
    {  
        memset(h,0,sizeof(h));  
        for(int i=1;i<=n;i++)scanf("%s",a[i]+1);  
        X[0]=1;  
        for(int i=1;i<=n;i++)  
        {  
            for(int j=1;j<=m;j++)  
            {  
                h[i][j]=h[i][j-1]+X[(i-1)*m+j-1]*(a[i][j]-'a');  
                X[(i-1)*m+j]=X[(i-1)*m+j-1]*19260817;  
            }  
        }  
        for(int j=1;j<=m;j++)  
        {  
            for(int i=1;i<=n;i++)  
            {  
                h[i][j]=h[i][j]+h[i-1][j];  
            }  
        }  
        int ans=0;  
        int l=0;  
        int r=min(n,m);  
        while(r-l>=0)  
        {  
            int mid=(l+r)/2;  
            if(Slove(mid)==1)  
            {  
                ans=mid;  
                l=mid+1;  
            }  
            else r=mid-1;  
        }  
        printf("%d\n",ans);  
    }  
}