易错点：
- 仔细思考可以发现，对角线上都有黑点等价于每一行都有起码一列与之对应。因此，对于每个黑点，将黑点所在行与所在列连边，进行二分图最大匹配即可.
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int MAXN=40010,MAXM=1000010;
struct Edge{
int from,to,nxt;
}e[MAXM];
int head[MAXN],edgeCnt=1;
void addEdge(int u,int v){
e[++edgeCnt].from=u;
e[edgeCnt].to=v;
e[edgeCnt].nxt=head[u];
head[u]=edgeCnt;
}
bool vis[MAXN];
int match[MAXN];
bool dfs(int x){
for(int i=head[x];i;i=e[i].nxt){
int nowV=e[i].to;
if(vis[nowV])continue;
vis[nowV]=1;
if(!match[nowV]||dfs(match[nowV])){
match[nowV]=x;
return 1;
}
}
return 0;
}
void init(){
memset(match,0,sizeof(match));
memset(e,0,sizeof(e));
edgeCnt=1;
memset(head,0,sizeof(head));
}
int main(){
int t;
scanf("%d",&t);
while(t--){
init();
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++){
int tmp;
scanf("%d",&tmp);
if(tmp)addEdge(i,j);
}
int ans=0;
for(int i=1;i<=n;i++){
memset(vis,0,sizeof(vis));
if(dfs(i))ans++;
}
if(ans>=n)printf("Yes
");
else printf("No
");
}
return 0;
}