【训练】2017-11-6晚-优快云博客

本文分享了三道竞赛编程题目的解题思路与实现方法，包括二分图匹配、找规律及最小生成树问题，提供了清晰的代码示例。
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T1

T1的话是一个明显得不能再明显得二分图匹配，然而我少判断了一组情况80P
#include<map>
#include<queue>
#include<cmath>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define qread(x) x=read()
#define mes(x,y) memset(x,y,sizeof(x))
#define mpy(x,y) memcpy(x,y,sizeof(x))
#define Maxn 1000
#define Maxm 100000
#define INF 2147483647  
inline int read(){
    char ch=getchar();
    int f=1,x=0;
    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

using namespace std;
struct node{
    int x,y,next;
}a[Maxm+1];int len,first[Maxn+1];
int t,n,m,k,ans,xlnum,ylnum,match[Maxn+1],v[Maxn+1];
bool xnum[Maxn+1],ynum[Maxn+1];
void ins(int x,int y){
    len++;
    a[len].x=x;a[len].y=y;
    a[len].next=first[x];first[x]=len; 
}
bool findmuniu(int x){
    for(int k=first[x];k>0;k=a[k].next){
        int y=a[k].y;
        if(v[y]!=v[0]){
            v[y]=v[0];
            if(match[y]==0||findmuniu(match[y])==true){
                match[y]=x;
                return true;
            }
        }
    }
    return false;
}
int main(){
	qread(t);
	while(t--){
		qread(n);qread(m);qread(k);
        len=0;mes(first,0);
        xlnum=0;mes(xnum,false);
        ylnum=0;mes(ynum,false);
        for(int i=1;i<=k;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            if(xnum[x]==false){
            	xnum[x]=true;
            	xlnum++;
            }
            if(ynum[y]==false){
            	ynum[y]=true;
            	ylnum++;
            }
            ins(x,y);
        }
        mes(match,0);
        ans=0;mes(v,0);
        for(int i=1;i<=n;i++){
            v[0]++;
            if(findmuniu(i)==true){
                ans++;
            }
        }
        if(ans==xlnum){
        	if(ans==ylnum)printf("Bob Win!\n");
        	else printf("Alice Win!\n");
        }
        else{
        	if(ans==ylnum)printf("Alice Win!\n");
        	else printf("Bob Win!\n");
        }
    }
    return 0;
}

下面的才是正解
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int Maxn = 1010;
const int Maxm = 100010;
struct node {
	int y, next;
}a[Maxm]; int first[Maxn], len;
void ins(int x, int y) {
	len++;
	a[len].y = y;
	a[len].next = first[x]; first[x] = len;
}
struct lnode {
	int x, y;
}list[Maxm];
int b[Maxm], c[Maxm], bl, cl;
int n, m, K;
int T, v[Maxn], belong[Maxn];
bool fc(int x) {
	for(int k = first[x]; k; k = a[k].next){
		int y = a[k].y;
		if(v[y] != T){
			v[y] = T;
			if(!belong[y] || fc(belong[y])){ belong[y] = x; return true; }
		}
	}
	return false;
}
int main() {
	freopen("a.in", "r", stdin);
	freopen("a.out", "w", stdout);
	int i, j, k;
	int Ti;
	scanf("%d", &Ti);
	while(Ti--){
		scanf("%d%d%d", &n, &m, &K);
		for(i = 1; i <= K; i++){
			scanf("%d%d", &list[i].x, &list[i].y);
			b[i] = list[i].x;
			c[i] = list[i].y;
		}
		sort(b+1, b+K+1);
		bl = unique(b+1, b+K+1) - (b+1);
		sort(c+1, c+K+1);
		cl = unique(c+1, c+K+1) - (c+1);
		if(bl != cl){ printf("Alice Win!\n"); continue; }
		len = 0; memset(first, 0, sizeof(first));
		for(i = 1; i <= K; i++){
			int x = lower_bound(b+1, b+bl+1, list[i].x) - b;
			int y = lower_bound(c+1, c+cl+1, list[i].y) - c;
			ins(x, y);
		}
		bool bk = true;
		memset(v, 0, sizeof(v));
		memset(belong, 0, sizeof(belong));
		for(T = 1; T <= bl; T++){
			if(!fc(T)){ bk = false; break; }
		}
		if(bk == true) printf("Bob Win!\n");
		else printf("Alice Win!\n");
	}
	return 0;
}
然后好好比对了一下？wak，粗心看错了条件。

T2

T2的话是一题找规律，然而我不会，所以我还是会给std

然而std时用dp做的？？（用dp找规律）
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;
const LL Maxn = 1100;
LL f[210][Maxn];
LL na[Maxn];
LL pr[210], prl;
LL n;
bool v[Maxn];
LL get_pr (){
    memset ( v, false, sizeof (v) );
    LL i, j;
    prl = 0;
    for ( i = 2; i <= 1000; i ++ ){
        if ( v[i] == false ){
            pr[++prl] = i;
            na[i] = prl;
        }
        for ( j = 1; j <= prl && i*pr[j] <= 1000; j ++ ){
            v[i*pr[j]] = true;
            if ( i % pr[j] == 0 ) break;
        }
    }
}
int main (){
	freopen("b.in", "r", stdin);
	freopen("b.out", "w", stdout);
    LL i, j, k;
    scanf ( "%lld", &n );
    get_pr ();
    memset ( f, 0, sizeof (f) );
    f[0][0] = 1;
    for ( i = 1; i <= prl; i ++ ){
        for ( j = 0; j <= n; j ++ ){
            f[i][j] += f[i-1][j];
            for ( k = pr[i]; k <= j; k *= pr[i] ){
                f[i][j] += f[i-1][j-k];
            }
        }
    }
    LL ans = 0;
    for ( i = 0; i <= n; i ++ ) ans += f[prl][i];
    printf ( "%lld\n", ans );
    return 0;
}
T3

题目就是要求最小生成树啊，然后就从每个点开始bfs，记录每个格子是谁先到的，后面的人再到达就和他连一条边，再做最小生成树就很棒了

然而我还是不会。所以送上一个90P的vio
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#pragma comment(linker, "/STACK:102400000,102400000") 
using namespace std;
const int dx[4]={0,-1,0,1};
const int dy[4]={-1,0,1,0};
struct ac
{
	int x,y,c,next;
}e[410000];int tlen,last[210000];
void insx(int x,int y,int c){tlen++;e[tlen].x=x;e[tlen].y=y;e[tlen].c=c;e[tlen].next=last[x];last[x]=tlen;}
struct node
{
	int x,y,dep,op;
}list[11100000];int head,tail;
struct edge
{
	int x,y,c;
}a[16160400];int len;
void ins(int x,int y,int c)
{
	len++;
	a[len].x=x;a[len].y=y;a[len].c=c;
}
int v[2100][2100],d[2100][2100];
bool cmp(edge n1,edge n2){return n1.c<n2.c;}
int h,w,p,q;
bool vis[2100][2100];
char s[2100][2100];
int fa[210000];
int findfa(int x)
{
	if(fa[x]!=x)fa[x]=findfa(fa[x]);
	return fa[x];
}
int f[210000][22],dep[810000];
int maxn[210000][22];
void pre_tree_node(int x)
{
	for(int i=1;i<=18;i++)f[x][i]=f[f[x][i-1]][i-1],maxn[x][i]=max(maxn[x][i-1],maxn[f[x][i-1]][i-1]);
	for(int k=last[x];k;k=e[k].next)
	{
		int y=e[k].y;
		if(y!=f[x][0])
		{
			f[y][0]=x;maxn[y][0]=e[k].c;
			dep[y]=dep[x]+1;
		//	printf("%d %d %d\n",e[k].x,e[k].y,e[k].c);
			pre_tree_node(y);
		}
	}
}
int getsum(int x,int y)
{
	int ret=0;
	if(dep[x]<dep[y])swap(x,y);
	for(int i=17;i>=0;i--)
		if(dep[f[x][i]]>=dep[y])ret=max(ret,maxn[x][i]),x=f[x][i];
	if(x==y)return ret;
	for(int i=17;i>=0;i--)
		if(f[x][i]!=f[y][i])ret=max(ret,max(maxn[x][i],maxn[y][i])),x=f[x][i],y=f[y][i];
	return max(ret,max(maxn[x][0],maxn[y][0]));
}
int main()
{
	scanf("%d%d%d%d",&h,&w,&p,&q);
	for(int i=1;i<=h;i++)scanf("%s",s[i]+1);
	memset(vis,true,sizeof(vis));
	memset(v,0,sizeof(v));
	head=1;tail=0;
	for(int i=1;i<=p;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		list[++tail].x=x;list[tail].y=y;list[tail].dep=0;list[tail].op=i;
		vis[x][y]=false;v[x][y]=i;d[x][y]=0;
	}
	while(head<=tail)
	{
		node tno=list[head];
		for(int i=0;i<=3;i++)
		{
			node next=tno;next.x+=dx[i];next.y+=dy[i];
			if(next.x>=1 && next.x<=h && next.y>=1 && next.y<=w)
			{
				if(vis[next.x][next.y]==true && s[next.x][next.y]!='#')
				{
					tail++;
					list[tail]=next;list[tail].dep=next.dep+1;
					v[next.x][next.y]=list[tail].op;
					d[next.x][next.y]=list[tail].dep;
					vis[next.x][next.y]=false;
				}
				else if(vis[next.x][next.y]==false && s[next.x][next.y]!='#' && v[next.x][next.y]!=v[tno.x][tno.y])
				{
					ins(v[next.x][next.y],v[tno.x][tno.y],d[next.x][next.y]+d[tno.x][tno.y]);
				}
			}
		}
		head++;
	}
	sort(a+1,a+1+len,cmp);
	for(int i=1;i<=p;i++)fa[i]=i;
	int cnt=p;tlen=0;memset(last,0,sizeof(last));
	for(int i=1;i<=len;i++)
	{
		int u=findfa(a[i].x),v=findfa(a[i].y);
		if(u!=v)
		{
			fa[u]=v;
			insx(a[i].x,a[i].y,a[i].c);
			insx(a[i].y,a[i].x,a[i].c);
			cnt--;
			if(cnt==1)break;
		}
	}
//	for(int i=1;i<=tlen;i+=2)printf("%d %d %d\n",e[i].x,e[i].y,e[i].c);
	for(int i=1;i<=p;i++)
		if(fa[i]==i)
		{
			dep[i]=1;pre_tree_node(i);
		}
/*	else
	{
		for(int i=1;i<=p;i++)
			if(fa[i]==i)
			{
				f[i]=0;dep[i]=0;dep_s[1]=0;
				pre_tree_node(i);
			}
	}*/
	while(q--)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		int u=findfa(x),v=findfa(y);
		if(u!=v){printf("-1\n");continue;}
		printf("%d\n",getsum(x,y)); 
	}
	return 0;
}
 

然后再送上一个100p的std
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
using namespace std;
const int Maxn = 2010;
const int Maxp = 200010;
const int lg = 18;
const int dx[4] = {0, -1, 0, 1};
const int dy[4] = {1, 0, -1, 0};
struct Point {
    int x, y;
};
struct enode {
    int x, y, d;
}list[Maxn*Maxn*4]; int l;
bool cmp(enode x, enode y) { return x.d < y.d; }
struct node {
    int y, next, d;
}a[Maxp*2]; int first[Maxp], len;
void ins(int x, int y, int d) {
    len++;
    a[len].y = y; a[len].d = d;
    a[len].next = first[x]; first[x] = len;
}
queue <Point> q;
int n, m, P, Q;
char s[Maxn][Maxn];
int num[Maxn][Maxn], dis[Maxn][Maxn];
int faa[Maxp];
int ff(int x) { return faa[x] == x ? x : faa[x] = ff(faa[x]); }
int fa[Maxp][lg], Max[Maxp][lg], dep[Maxp];
int _max(int x, int y) { return x > y ? x : y; }
void dfs(int x) {
    for(int i = 1; i < lg; i++) fa[x][i] = fa[fa[x][i-1]][i-1], Max[x][i] = _max(Max[x][i-1], Max[fa[x][i-1]][i-1]);
    for(int k = first[x]; k; k = a[k].next){
        int y = a[k].y;
        if(y == fa[x][0]) continue;
        dep[y] = dep[x]+1; fa[y][0] = x; Max[y][0] = a[k].d;
        dfs(y);
    }
}
int getlca(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    int ret = 0;
    for(int i = lg-1; i >= 0; i--){
        if(dep[fa[x][i]] >= dep[y]) ret = _max(ret, Max[x][i]), x = fa[x][i];
    }
    if(x == y) return ret;
    for(int i = lg-1; i >= 0; i--){
        if(fa[x][i] != fa[y][i]) ret = _max(ret, _max(Max[x][i], Max[y][i])), x = fa[x][i], y = fa[y][i];
    }
    return _max(ret, _max(Max[x][0], Max[y][0]));
}
int main() {
	freopen("c.in", "r", stdin);
	freopen("c.out", "w", stdout);
    int i, j, k;
    scanf("%d%d%d%d", &n, &m, &P, &Q);
    for(i = 1; i <= n; i++) scanf("%s", s[i]+1);
    for(i = 1; i <= P; i++){
        Point o;
        scanf("%d%d", &o.x, &o.y);
        num[o.x][o.y] = i; dis[o.x][o.y] = 0;
        q.push(o);
    }
    while(!q.empty()){
        Point x = q.front(); q.pop();
        for(k = 0; k < 4; k++){
            Point y;
            y.x = x.x+dx[k]; y.y = x.y+dy[k];
            if(y.x < 1 || y.x > n || y.y < 1 || y.y > m || s[y.x][y.y] == '#') continue;
            if(!num[y.x][y.y]){
                num[y.x][y.y] = num[x.x][x.y]; dis[y.x][y.y] = dis[x.x][x.y]+1;
                q.push(y);
            } else if(num[x.x][x.y] != num[y.x][y.y]){
                l++;
                list[l].x = num[x.x][x.y]; list[l].y = num[y.x][y.y]; list[l].d = dis[x.x][x.y]+dis[y.x][y.y];
            }
        }
    }
    for(i = 1; i <= P; i++) faa[i] = i;
    sort(list+1, list+l+1, cmp);
    for(i = 1; i <= l; i++){
        int fx = ff(list[i].x), fy = ff(list[i].y);
        if(fx != fy){
            faa[fx] = fy;
            ins(list[i].x, list[i].y, list[i].d);
            ins(list[i].y, list[i].x, list[i].d);
        }
    }
    for(i = 1; i <= P; i++){
        if(faa[i] == i) dep[i] = 1, dfs(i);
    }
    for(i = 1; i <= Q; i++){
        int x, y;
        scanf("%d%d", &x, &y);
        int fx = ff(x), fy = ff(y);
        if(fx != fy){ printf("-1\n"); continue; }
        printf("%d\n", getlca(x, y));
    }
    return 0;
}
 

查看原文：http://hz2016.cn/blog/?p=115