链接:点击打开链接
题意:有F种食物和D种饮料,每种食物或饮料只能供有限次,且每个人只享用一种食物和一种饮料。现在有n头牛,每个人都有自己喜欢的食物种类和饮料的种类,问最多能使几个人同时享用到自己喜欢的食物和饮料
代码:
#include <queue> #include <vector> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> using namespace std; const int INF=0x3f3f3f3f; struct node{ int u,v,cap; node(){} node(int u,int v,int cap):u(u),v(v),cap(cap){} }es[505*505]; int R,S,T; int dis[5005],iter[5005]; vector<int> tab[5005]; void addedge(int u, int v, int cap){ tab[u].push_back(R); es[R++]=node(u,v,cap); tab[v].push_back(R); es[R++]=node(v,u,0); } int bfs(){ int i,h; queue<int> q; q.push(S); memset(dis,INF,sizeof(dis)); dis[S]=0; while(q.size()){ h=q.front(); q.pop(); for(i=0;i<tab[h].size();i++){ node &e=es[tab[h][i]]; if(e.cap>0&&dis[e.v]==INF){ dis[e.v]=dis[h]+1; q.push(e.v); } } } return dis[T]<INF; } int dfs(int x,int maxflow){ int flow; if(x==T) return maxflow; for(int &i=iter[x];i<tab[x].size();i++){ node &e=es[tab[x][i]]; if(dis[e.v]==dis[x]+1&&e.cap>0){ flow=dfs(e.v,min(maxflow,e.cap)); if(flow){ e.cap-=flow; es[tab[x][i]^1].cap+=flow; return flow; } } } return 0; } int dinic(){ int ans,flow; ans=0; while(bfs()){ memset(iter,0,sizeof(iter)); while(flow=dfs(S,INF)) ans+=flow; } return ans; } //dinic模板 int main(){ //与poj3281相同 char s[5005]; //建图为: int N,F,D,i,j,num; //S->食物->人->人->饮料->T while(scanf("%d%d%d",&N,&F,&D)!=EOF){ S=0,R=0,T=F+2*N+D+1; for(i=0;i<=T;i++) tab[i].clear(); for(i=1;i<=F;i++){ scanf("%d",&num); addedge(S,i,num); } for(i=F+2*N+1;i<=F+2*N+D;i++){ scanf("%d",&num); addedge(i,T,num); } for(i=F+1;i<=F+N;i++) addedge(i,i+N,1); for(i=1;i<=N;i++){ scanf("%s",s); for(j=0;j<F;j++){ if(s[j]=='Y') addedge(j+1,F+i,1); } } for(i=1;i<=N;i++){ scanf("%s",s); for(j=0;j<D;j++){ if(s[j]=='Y') addedge(F+N+i,F+2*N+j+1,1); } } printf("%d\n",dinic()); } return 0; }