[luogu1129 ZJOI2007] 矩阵游戏 (二分图最大匹配)

传送门

Description

Input

Output

Sample Input

2
2
0 0
0 1
3
0 0 1
0 1 0
1 0 0

Sample Output

No
Yes

HINT

Solution

经简单证明可知原题等价于给你一些黑色格子，问能否选出n个，使得每行、每列有且仅有一个黑色格子
那么直接对黑色格子的行列连边跑最小点覆盖

Code

//By Menteur_Hxy
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define F(i,a,b) for(register int i=(a);i<=(b);i++)
#define E(i,u) for(register int i=head[u];i;i=nxt[i])
#define ins(a,b,c) add(a,b,c),add(b,a,0)
#define add(a,b,c) nxt[++cnt]=head[a],to[cnt]=b,cst[cnt]=c,head[a]=cnt
using namespace std;

int read() {
	int x=0,f=1; char c=getchar();
	while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
	while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
	return x*f;
}

const int N=2010,INF=0x3f3f3f3f;
int n,S,T,cnt;
int nxt[N*N],to[N*N],cst[N*N],head[N<<1],cur[N<<1];
int que[N<<1],dis[N<<1];

bool bfs() {
	memset(dis,0,sizeof(dis));
	int h=0,t=1; que[++h]=S; dis[S]=1;
	while(h<=t) {
		int u=que[h++];
		// cout<<u<<" "<<dis[u]<<endl;
		E(i,u) if(!dis[to[i]]&&cst[i]>0) 
			dis[to[i]]=dis[u]+1,que[++t]=to[i];
	}
	return dis[T];
}

int dfs(int u,int flow) {
	if(u==T) return flow;
	int used=0;
	for(register int &i=cur[u],v;i;i=nxt[i]) {
		// cout<<u<<" "<<to[i]<<" "<<cst[i]<<" "<<dis[to[i]]<<" "<<dis[u]<<endl;
		if(cst[i]>0&&dis[(v=to[i])]==dis[u]+1) {
			int tmp=dfs(v,min(flow-used,cst[i]));
			cst[i]-=tmp,cst[i^1]+=tmp;
			used+=tmp; if(used==flow) return flow;
		}
	}
		
	if(!used) dis[u]=-1;
	return used;
}

int dinic() {
	int res=0;
	while(bfs()) {
		F(i,0,T) cur[i]=head[i];
		res+=dfs(S,INF);
	}
	return res;
}

int main() {
	int cas=read();
	while(cas--) {
		n=read(); S=0,T=(n<<1|1),cnt=1;
		memset(head,0,sizeof(head));
		F(i,1,n) ins(S,i,1); F(i,1,n) ins(i+n,T,1);
		F(i,1,n) F(j,1,n) if(read()) ins(i,j+n,1);
		int res=dinic();
		if(res==n) puts("Yes");
		else puts("No");
	}
	return 0;
}