[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;

}