常见问题五大算法思想（一）分治算法及常见例子_介绍分治法ppt-优快云博客
芯片检测问题
大数相乘
最后return公式推导详见分治法的经典问题——大整数相乘 - m0w3n - 博客园 (cnblogs.com)
#include <iostream>
using namespace std;
#include <cmath>
/* num1：第一个乘数
* num2：第二个乘数
* num1length：第一个乘数的个数
* num2length：第二个乘数的个数
$$ W ( n ) = 3 W ( n / 2 ) + c n$$ ++--工作量
*/
long bigIntPow(long num1, long num2, int num1length, int num2length){
if (num1 == 0 || num2 == 0){
return 0;
}else if (num1length == 1 || num2length == 1){
return num1*num2;
}else{
int xn0 = num1length / 2, yn0 = num2length / 2;
int xn1 = num1length - xn0, yn1 = num2length - yn0;
//123 -> 123/10 = 12, 123%10 = 3
long A = (long)(num1 / pow(10, xn0));
long B = num1 % (long long)pow(10, xn0);
long C = (long)(num2 / pow(10, yn0));
long D = num2 % (long long)pow(10, yn0);
long AC = bigIntPow(A, C, xn1, yn1);
long BD = bigIntPow(B, D, xn0, yn0);
//AD+BC=(A-B)*(D-C)……
long ABCD = bigIntPow((long)(A * pow(10, xn0) - B), (long)(D - C * pow(10, yn0)), xn1, yn1);
return (long)(2 * AC *pow(10, (xn0 + yn0)) + ABCD + 2 * BD);
}
}
int main() {
int num1 = 785;
int num2 = 123;
int num1length = 3;
int num2length = 3;
long result = bigIntPow(num1,num2,num1length,num2length);
cout<<"普通乘法 X*Y="<<num1<<"*"<<num2<<"="<<num1*num2<<endl;
cout<<"分治乘法 X*Y="<<num1<<"*"<<num2<<"="<<result<<endl;
}
平面最近点对
/*
复杂度分析：
步1递归边界处理： 0(1)
步2排序： O(nlogn)
步3划分： 0(1)
步4-5子问题： 2T(n/2)
步6确定8： 0(1)
步7检查跨边界点对：0(n)
*/
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
#include<iostream>
using namespace std;
bool same(double a, double b) { /// 1e-5精度意义下的浮点数相等
if(fabs(a-b) <= 1e-5) return true; return false;
}
struct node{
double x,y;
};
node arr[1000];
double dist(node x1,node x2){
return sqrt((x1.x-x2.x)*(x1.x-x2.x)+(x1.y-x2.y)*(x1.y-x2.y));
}
bool cmpx(node a,node b){
return a.x<b.x;
}
bool cmpy(node a,node b){
return a.y<b.y;
}
double mind(int L,int R){
sort(arr+L,arr+R+1,cmpx);
double ans=1e9;
if(R-L+1<=3){
for(int i=L;i<=R;i++)//1 - 3
for(int j=i+1;j<=R;j++)//2 - 3
ans=min(ans,dist(arr[i],arr[j]));
}
else{
int mid=(L+R)/2;//第中间的那个划分
double midx=arr[mid].x;//第中间的那个的 x值
ans=min(ans,mind(L,mid));
ans=min(ans,mind(mid+1,R));//d
vector<node> avai; avai.clear();
// 用vector存一下“竖条”范围中的点
//遍历点找出与竖条距离最近的点集合
for(int i=L;i<=R;i++)
if(abs(arr[i].x-midx)<ans)
avai.push_back(arr[i]);
double dnow=1e9;
sort(avai.begin(),avai.end(),cmpy);
for(int i=0;i<avai.size();i++){
for(int j=i+1; j<avai.size();j++){
double d=dist(avai[i],avai[j]);
if(d>ans && !same(d,ans)) break;//再遍历会越来越大
//就是一个点 配对下一个点（这是两点间的最近距离），大于了当前的最小距离，跳出
dnow=min(dnow,d);
}
}
ans=min(ans,dnow); avai.clear();
}
return ans;
}
int main(){
int n;
cin>>n;
for(int i=1;i<=n;i++) cin>>arr[i].x>>arr[i].y;
double ans=mind(1,n);
cout<<ans;
}
优化版平面最近点对的分治做法及其证明_第二题(较难,选做题,做出来的加3分)题目描述给定平面上n个点,找出其中的一对-优快云博客
找出最大最小值FindMaxMin
分组算法(分成两组）
将n个元素两两一组分成Ln/2」组
每组比较，得到Ln/2」个较小和Ln/2」个较大
在「n/2个(n为奇数，是Ln/2」+1)较小中找最小min
在「n/2个(n为奇数，是Ln/2」+1)较大中找最大max
//算法2.9 FindMaxMin
//输入：n个数的数组
//输出：max，min
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
using namespace std;
void show(int array[], int len)
{
int i;
for(i = 0; i < len; i++)
{
printf("%-3d ", array[i]);
}
printf("\n");
return ;
}
//找最大
void Findmax(int L[],int len){
int max,i;
max = L[0];
for (i = 0; i < len; i++){
if (max < L[i]){
max = L[i];
}
}
printf("最大值为%d\n", max);
}
//找最小
void Findmin(int L[],int len){
int min,i;
min = L[0];
for (i = 0; i < len; i++){
if (min > L[i]){
min = L[i];
}
}
printf("最小值为%d\n", min);
}
//分组
void ass(int arr[],int p,int r,int a[])
{
int i,j=0;
for(i=p;i<=r;i++)
{
a[j++]=arr[i];
cout<<arr[i]<<" ";
}puts("");
}
void FindMaxMin(int L[],int r)
{
int *left,*right,i;
int mid=(r+1)/2;
printf("mid=%d\n",mid);
if((r+1)%2!=0) //奇数
{ cout<<mid<<endl;
ass(L,0,mid-1,left);
ass(L,mid+1,r,right);
}else{ //偶数
ass(L,0,mid-1,left);
ass(L,mid,r,right);
}
//较小的值都被放在 left 数组中，而较大的值都被放在 right 数组中
for(i=0;i<mid;i++)
{
int temp;
if(left[i]>right[i])
{
temp=left[i];
left[i]=right[i];
right[i]=temp;
}
}
if((r+1)%2!=0) { //奇数
//加入落单的元素
left[mid]=L[mid];
right[mid]=L[mid];
printf("left[]=");
show(left,mid+1);
printf("right[]=");
show(right,mid+1);
Findmin(left,mid+1);
Findmax(right,mid+1);
}else{ //偶数
printf("left[]=:");
show(left,mid);
printf("right[]=") ;
show(right,mid);
Findmin(left,mid);
Findmax(right,mid);
}
}
int main(){
int L[] = {0,1,2,3,4};
int i, j,len;
len=sizeof(L)/sizeof(L[0]);
printf("数组个数为；%d\n",len);
show(L,len);
FindMaxMin(L,len-1);//-1
} 分治算法
#include<iostream>
#include<algorithm>
using namespace std;
//在数组a的区间[i,j]范围内寻找一个最大值和一个最小值并通过指针*max和*min返回
int findMaxMin(int *a,int i,int j,int *max,int *min){
if(i==j){
*max = a[i];
*min = a[i];
return 0;
}else if(i+1==j){
*max = std::max(a[i],a[j]);
*min = std::min(a[i],a[j]);
return 0;
}
int mid;
int lmax,lmin,rmax,rmin;
mid = i+(j-i)/2;
findMaxMin(a,i,mid,&lmax,&lmin);
findMaxMin(a,mid+1,j,&rmax,&rmin);
*max = std::max(lmax,rmax);
*min = std::min(lmin,rmin);
return 0;
}
int main(){
int S[]= {1,3,5,7,9,2,4,6,8,10};//10
int n = sizeof(S)/sizeof(int);
int max,min;
findMaxMin(S,0,n-1,&max,&min);
cout << "min:" << min << endl;
cout << "max:" << max << endl;
}第k大的元素Learning Select Algorithms - Hank's Blog (zhengyhn.github.io)
#include<iostream>
#include<algorithm>
using namespace std;
void insertSort(int R[], int low, int high) {//开始索引结束索引
int i, j, tmp;
for (i = low + 1; i <= high; ++i) {
tmp = R[i]; //假设i指向第二个元素
j = i - 1; //j指向第一个元素
while (j >= low && R[j] > tmp) { //假设第一个元素>第二个元素
R[j + 1] = R[j]; //第一个元素右移一位，此时第二个位置上为较大的数
--j; //j指向下一个（前一个）
}
R[j + 1] = tmp; //把最后一次-1加回来，把第二个元素放在第一个位置上。
}
}
int FindMid(int R[], int low, int high) {//*mid 中值 分组n/5
if (low == high)return R[low];
int i, k;
for (i = low; i + 4 <= high; i += 5) {// 01234 56789 134 三组[0,1]
insertSort(R, i, i + 4); // i i 分组排序
k = i - low;// 5-0=5
//第一组的数据，R[i + 2]：第i组的中位数
//将每组的中位数移动到数组开头
swap(R[low + k / 5], R[i + 2]); // 0+5/5=1 i=5+2=7
}
int n = high - i + 1; //12-10+1=3 不是5个完整组的开头的个数
if (n > 0) {
insertSort(R, i, high);
k = i - low;//初始位置-low=10-0=10
swap(R[low + k / 5], R[i + n / 2]);
}
k = k / 5;//2
if (k == 0) return R[low];
return FindMid(R, low, low + k);//下一轮 0+2-1
}
int FindId(int R[], int low, int high, int median) {
for (int i = low; i <= high; ++i) {
if (median == R[i]) {
return i;
}
}
return -1;
}
int Partion(int R[], int low, int high, int index) { //以R[index]为枢纽值，做一次划分操作，左边比它小，右边比它大
if (low <= high) {
swap(R[index], R[low]);
/*
12345
32145
*/
int tmp = R[low];//枢纽值3
int i = low, j = high;
while (i != j) {
while (j > i&& R[j] >= tmp) {
--j;
}
R[i] = R[j];//3 5 52145
while (i < j && R[i] <= tmp) {
++i;
}
R[j] = R[i];//
}
R[i] = tmp;
return i;
}
return -1;
}
int Select(int R[], int low, int high, int k) {
int median = FindMid(R, low, high);
int index = FindId(R, low, high, median);//找到最优解m的索引
int newIndex = Partion(R, low, high, index);//找到m在排序数组的索引
int rank = newIndex - low + 1;// 中间第rank大的元素
if (rank == k) return R[newIndex];
else if (rank > k) return Select(R, low, newIndex - 1, k);//比rank小
return Select(R, newIndex + 1, high, k - rank);
}
int main()
{
int i, n, k;
int d[] = { 8,2,3,5,7, 6,11,14,1,9, 13,10,4,12,15 };
n = sizeof(d) / sizeof(int);
k = 6;
int index;
index = Select(d, 0, n - 1, k);
cout << "第" << k << "大的元素是" << index << endl;
return 0;
}纯享版
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
int select5(std::vector<int>& arr, int left, int right);
int get_pivot(std::vector<int>& arr, int left, int right);
int select(std::vector<int>& arr, int left, int right, int k);
int bfprt_select(std::vector<int>& arr, int k);
int select5(std::vector<int>& arr, int left, int right) {
std::sort(arr.begin() + left, arr.begin() + right + 1);
return left + (right - left) / 2;
}
int get_pivot(std::vector<int>& arr, int left, int right) {
if (right - left < 5) {
return select5(arr, left, right);
}
for (int i = left; i < right - 4; i += 5) {
int sub_right = i + 4;
if (sub_right > right) {
sub_right = right;
}
int median_index = select5(arr, i, sub_right);
std::swap(arr[median_index], arr[left + (i - left) / 5]);
}
return select(arr, left, left + (right - left) / 5, (right - left) / 10 + 1);
}
int select(std::vector<int>& arr, int left, int right, int k) {
while (true) {
if (left >= right) {
return left;
}
int pivot_index = get_pivot(arr, left, right);
std::swap(arr[right], arr[pivot_index]);
int store_index = left;
for (int i = left; i < right; i++) {
if (arr[i] < arr[right]) {
std::swap(arr[i], arr[store_index]);
store_index++;
}
}
std::swap(arr[right], arr[store_index]);
if (k - 1 == store_index) {
return store_index;
} else if (k - 1 > store_index) {
left = store_index + 1;
} else {
right = store_index - 1;
}
}
}
int bfprt_select(std::vector<int>& arr, int k) {
int idx = select(arr, 0, arr.size() - 1, k);
return arr[idx];
}
int main() {
int n;
std::cout << "Enter the size of the array: ";
std::cin >> n;
std::vector<int> arr(n);
std::cout << "Enter the elements of the array: ";
for (int& i : arr) {
std::cin >> i;
}
int k;
std::cout << "Enter the kth smallest number to find: ";
std::cin >> k;
std::cout << "The " << k << "th smallest number is: " << bfprt_select(arr, k) << "\n";
return 0;
}
最大子序列Maximum SubArray Problem (最大子数组问题) 分治算法的改进 | Fancy's Blog (fancypei.github.io)
#include<iostream>
using namespace std;
int maxsubSum(int a[],int left,int right){
int sum=0;
if(left==right){return a[left];}
int center=(left+right)/2;
int leftsum=maxsubSum(a,left,center);
int rightsum=maxsubSum(a,center+1,right);
//从中间向两边散开
int s1=0,lefts=0;
for(int i=center;i>=left;i--){
lefts+=a[i];
if(lefts>s1) s1=lefts;
}
int s2=0,rights=0;
for(int i=center+1;i<=right;i++){
rights+=a[i];
if(rights>s2) s2=rights;
}
sum=max(leftsum,max(rightsum,s1+s2));
return sum;
}
int main(){
int a[] = {-2,11,-4,13,-5,-2,};
// left right
int n=sizeof(a)/sizeof(int);
for(int i= 0; i < 6; i++)
{
cout<<a[i]<<" ";
} cout<<endl;
cout<<"数组a的最大连续子段和为：" << maxsubSum(a,0,n-1)<<endl;
return 0;
}时间复杂度更低的版本
#include<iostream>
using namespace std;
/*
sum：段中所有元素的总和。
maxl：从l开始，到段内某个点结束的子段的最大和。
maxr：从段内某个点开始，到r结束的子段的最大和。
max：段内任何子段的最大和。
*/
struct node{
int sum, maxl, maxr, max;
};
node Max(int *a, int l, int r){
if(l == r){
return (node){a[l], a[l], a[l], a[l]};
}
int mid = (l + r) / 2;
node L = Max(a, l, mid);//递归调用段左半部分的Max
node R = Max(a, mid + 1, r);
node T;
T.sum = L.sum + R.sum;
T.maxl = max(L.maxl, L.sum + R.maxl);
T.maxr = max(R.maxr, R.sum + L.maxr);
T.max = max(L.max, max(R.max, L.maxr + R.maxl));
return T;
}
int main(){
int n, a[10001];
cin >> n;
for(int i = 1; i <= n; i++){
cin >> a[i];
}
cout << Max(a, 1, n).max << endl;
return 0;
}棋盘覆盖
#include <iostream>
#include <algorithm>
using namespace std;
/*
0 0 0 0
0 * 0 0
0 0 0 0
0 0 0 0
//划分后在左上角
(0,0) (1,1)
dr<tr+s
dc<tc+s
//如果没有就在次划分的方块右下角
(1,1)填充
*/
int t;
int num = 0;
int board[100][100];
void chessBoard(int tr, int tc, int dr, int dc, int size){
//(tr,tc) 左上角 (dr,dc) 特殊 size 为棋盘大小
if(size == 1){return;}//只有一个特殊方格
int t=++num;//L型骨牌号
// t++;
int s=size/2;
//划分后在左上角
if(dr<tr+s && dc<tc+s){
chessBoard(tr,tc,dr,dc,s);
}
else{//用t号骨牌填充右下角
board[tr+s-1][tc+s-1]=t;
chessBoard(tr,tc,tr+s-1,tc+s-1,s);
}
//划分后在右上角
if(dr<tr+s&&dc>=tc+s){
chessBoard(tr,tc+s,dr,dc,s);
}
else{//用t号骨牌填充左下角
board[tr+s-1][tc+s]=t;
chessBoard(tr,tc+s,tr+s-1,tc+s,s);//左上角新开始（tr，tc+s）
}
//划分后在左下角
if(dr>=tr+s&&dc<tc+s){
chessBoard(tr+s,tc,dr,dc,s);
}
else{//用t号骨牌填充右上角
board[tr+s][tc+s-1]=t;
chessBoard(tr+s,tc,tr+s,tc+s-1,s);
}
//划分后在右下角
if(dr>=tr+s&&dc>=tc+s){
chessBoard(tr+s,tc+s,dr,dc,s);
}
else{//用t号骨牌填充左上角
board[tr+s][tc+s]=t;
chessBoard(tr+s,tc+s,tr+s,tc+s,s);
}
}
int main()
{
int size,r,c,row,col;
printf("请输入棋盘的行列号");
scanf("%d",&size);
printf("请输入特殊方格的行列号");
scanf("%d %d",&row,&col);
chessBoard(0,0,row,col,size);
for (r = 0; r < size; r++)
{
for (c = 0; c < size; c++)
{
printf("%2d ",board[r][c]);
}
printf("\n");
}
return 0;
}
循环赛日程表
参考来源video-Algorithm-QWL/027findMaxMin2.cpp at master · JizhiXiang/video-Algorithm-QWL (github.com)
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