BZOJ 1059 矩阵游戏

注意到如果变换前在同一行/列，变换后也在同一行/列。然后二分图匹配。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define maxn 250
#define maxv 550
#define maxe 1000050
#define inf 2000000000
using namespace std;
int tt,n,x,nume,g[maxv],dis[maxv],s,t;
queue <int> q;
struct edge
{
    int v,f,nxt;
}e[maxe];
void addedge(int u,int v,int f)
{
    e[++nume].v=v;e[nume].f=f;e[nume].nxt=g[u];g[u]=nume;
    e[++nume].v=u;e[nume].f=0;e[nume].nxt=g[v];g[v]=nume;
}
bool bfs()
{
    for (int i=s;i<=t;i++) dis[i]=inf;
    dis[s]=0;q.push(s);
    while (!q.empty())
    {
        int head=q.front();q.pop();
        for (int i=g[head];i;i=e[i].nxt)
        {
            int v=e[i].v;
            if ((dis[v]==inf) && (e[i].f))
            {
                dis[v]=dis[head]+1;
                q.push(v);
            }
        }
    }
    if (dis[t]==inf) return false;
    return true;
}
int dinic(int x,int low)
{
    if (x==t) return low;
    int ret=0;
    for (int i=g[x];low && i;i=e[i].nxt)
    {
        int v=e[i].v;
        if ((e[i].f) && (dis[v]==dis[x]+1))
        {
            int dd=dinic(v,min(low,e[i].f));
            ret+=dd;low-=dd;
            e[i].f-=dd;e[i^1].f+=dd;
        }
    }
    if (!ret) dis[x]=inf;
    return ret;
}
void work()
{
    nume=1;memset(g,0,sizeof(g));
    scanf("%d",&n);s=0;t=2*n+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++)
        {
            scanf("%d",&x);
            if (x) addedge(i,j+n,1);
        }
    int max_flow=0;
    while (bfs()) max_flow+=dinic(s,inf);
    if (max_flow==n) printf("Yes
");
    else printf("No
");
}
int main()
{
    scanf("%d",&tt);
    for (int i=1;i<=tt;i++)
        work();
    return 0;
}