【BZOJ】1711: [Usaco2007 Open]Dining吃饭

【算法】最大流

【题解】

S连向食物连向牛连向牛‘连向饮料连向T。

经典的一个元素依赖于两个元素的建图方式。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=100010,inf=0x3f3f3f3f;
struct edge{int v,flow,from;}e[maxn];
int first[maxn],d[maxn],cur[maxn],tot=1,S,T,n,A,B;
void insert(int u,int v,int w){//printf("%d %d\n",u,v);tot++;e[tot].v=v;e[tot].flow=w;e[tot].from=first[u];first[u]=tot;tot++;e[tot].v=u;e[tot].flow=0;e[tot].from=first[v];first[v]=tot;
}
queue<int>q;
bool bfs(){memset(d,-1,sizeof(d));d[S]=0;q.push(S);while(!q.empty()){int x=q.front();q.pop();for(int i=first[x];i;i=e[i].from)if(e[i].flow&&d[e[i].v]==-1){d[e[i].v]=d[x]+1;q.push(e[i].v);}}return d[T]!=-1;
}
int dinic(int x,int a){if(x==T||a==0)return a;int flow=0,f;for(int& i=cur[x];i;i=e[i].from)if(e[i].flow&&d[e[i].v]==d[x]+1&&(f=dinic(e[i].v,min(e[i].flow,a)))>0){e[i].flow-=f;e[i^1].flow+=f;a-=f;flow+=f;if(a==0)break;}return flow;
}
int main(){scanf("%d%d%d",&n,&A,&B);S=0;T=A+n+n+B+1;int u,v,w;for(int i=1;i<=A;i++)insert(S,i,1);for(int i=1;i<=B;i++)insert(A+n+n+i,T,1);for(int i=1;i<=n;i++){scanf("%d%d",&u,&v);for(int j=1;j<=u;j++){scanf("%d",&w);insert(w,A+i,1);}for(int j=1;j<=v;j++){scanf("%d",&w);insert(A+n+i,A+n+n+w,1);}insert(A+i,A+n+i,1);}int ans=0;while(bfs()){for(int i=S;i<=T;i++)cur[i]=first[i];ans+=dinic(S,inf);}printf("%d",ans);return 0;
}

View Code

转载于:https://www.cnblogs.com/onioncyc/p/7450535.html

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