BZOJ 1059 [ZJOI2007]矩阵游戏

Posted 2020-10-25 ws_zzy

题解：二分图匹配

行列建点

每一条边i->j表示第i行移动到第j行符合题意，即a[i][j]=1

可证明不需考虑列的交换

求最大匹配即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int maxn=300;
const int oo=1000000000;

int T;
int n;

struct Edge{
	int from,to,cap,flow;
};
vector<int>G[maxn*maxn];
vector<Edge>edges;
void Addedge(int x,int y,int z){
	Edge e;
	e.from=x;e.to=y;e.cap=z;e.flow=0;
	edges.push_back(e);
	e.from=y;e.to=x;e.cap=0;e.flow=0;
	edges.push_back(e);
	int c=edges.size();
	G[x].push_back(c-2);
	G[y].push_back(c-1);
}

int s,t;
int vis[maxn*maxn];
int d[maxn*maxn];
queue<int>q;
int Bfs(){
	memset(vis,0,sizeof(vis));
	vis[s]=1;d[s]=0;q.push(s);
	while(!q.empty()){
		int x=q.front();q.pop();
		for(int i=0;i<G[x].size();++i){
			Edge e=edges[G[x][i]];
			if((e.cap>e.flow)&&(!vis[e.to])){
				vis[e.to]=1;
				d[e.to]=d[x]+1;
				q.push(e.to);
			}
		}
	}
	return vis[t];
}

int Dfs(int x,int a){
	if((x==t)||(a==0))return a;
	
	int nowflow=0,f;
	for(int i=0;i<G[x].size();++i){
		Edge e=edges[G[x][i]];
		if((d[x]+1==d[e.to])&&((f=Dfs(e.to,min(a,e.cap-e.flow)))>0)){
			nowflow+=f;
			a-=f;
			edges[G[x][i]].flow+=f;
			edges[G[x][i]^1].flow-=f;
			if(a==0)break;
		}
	}
	return nowflow;
}

int Maxflow(){
	int flow=0;
	while(Bfs())flow+=Dfs(s,oo);
	return flow;
}

void Dinicinit(){
	for(int i=0;i<=n+n+2;++i)G[i].clear();
	edges.clear();
	while(!q.empty())q.pop();
}

int main(){
	scanf("%d",&T);
	while(T--){
		Dinicinit();
		scanf("%d",&n);
		s=n+n+1;t=s+1;
		for(int i=1;i<=n;++i){
			Addedge(s,i,1);
			Addedge(i+n,t,1);
		}
		for(int i=1;i<=n;++i){
			for(int j=1;j<=n;++j){
				int x;
				scanf("%d",&x);
				if(x==1)Addedge(i,j+n,1);
			}
		}
		if(Maxflow()==n)printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}