二分图匹配——BZOJ1059/Luogu1129 [ZJOI2007]矩阵游戏

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http://www.lydsy.com/JudgeOnline/problem.php?id=1059
https://www.luogu.org/problem/show?pid=1129
我们可以很快的把这题转化成二分图匹配
如果匹配数小于n即输出no，否则输出yes
具体我不多说了。。。
然后直接上模板（我发现我以前写的是假的dinic）
具体见前一篇飞行员匹配：http://blog.csdn.net/jzq233jzq/article/details/64122738

#include<bits/stdc++.h>#define ll long longusing namespace std;ll dist[100001],nedge=-1,head[100001],nex[100001],p[100001],c[100001],m,n,s,t,x,y,z;inline void addedge(ll x,ll y,ll z){    nedge++;p[nedge]=y;    c[nedge]=z;nex[nedge]=head[x];    head[x]=nedge;}inline bool bfs(ll s,ll e){    ll now;    queue<ll>q;    memset(dist,-1,sizeof dist);    dist[s]=1;    q.push(s);    while(!q.empty()){        now=q.front();        q.pop();        ll k=head[now];        while(k>-1){            if(dist[p[k]]<0&&c[k]>0){                dist[p[k]]=dist[now]+1;                q.push(p[k]);            }            k=nex[k];        }    }    if(dist[e]>0)return 1;    return 0;}inline ll finds(ll x,ll low){    ll a;if(x==t)return low;    ll used=0,k=head[x];    while(k>-1){        if(c[k]&&dist[x]+1==dist[p[k]]&&(a=finds(p[k],min(low-used,c[k])))){            c[k]-=a;            c[k^1]+=a;            used+=a;        }        k=nex[k];        if(used==low)break;    }    if(used==0)dist[x]=-1;    return used;}inline ll dinic(ll s,ll e){    ll flow=0;    while(bfs(s,e))flow+=finds(s,0x3f3f3f3f);    return flow;}int main(){    scanf("%lld",&m);    while(m--){        nedge=-1;        memset(p,-1,sizeof p);        memset(c,-1,sizeof c);        memset(nex,-1,sizeof nex);        memset(head,-1,sizeof head);        scanf("%lld",&n);        for(ll i=1;i<=n;i++){            addedge(0,i,1);            addedge(i,0,0);        }        for(ll i=n+1;i<=n*2;i++){            addedge(i,2*n+1,1);            addedge(2*n+1,i,0);        }        for(ll i=1;i<=n;i++)            for(ll j=1;j<=n;j++){                int x;scanf("%d",&x);                if(!x)continue;                addedge(i,j+n,1);                addedge(j+n,i,0);            }        s=0;t=2*n+1;        if(dinic(s,t)<n)printf("No\n");        else printf("Yes\n");    }    return 0;}

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