HDU 4292 Food（网络流）

题意：有F种食物 D种饮料 它们都有一定的数量 有N个人 每个人都有自己喜欢吃的食物和饮料 （每个人至少要一种食物和饮料）只有能满足他的要求时他才会接服务 求最大能满足多少人？

思路：网络流 建一超级源点 汇点 源点与食物相连 边权为其数量，汇点与饮料相连 边权也为其数量 把人分成两个点 之间的边权为1每个人与之需要的食物和饮料相连 边权为1 

////125MS   2136K


#include <stdio.h>
#include <string.h>
#define VM   1000
#define EM   200000
#define inf 0x3f3f3f3f
struct E
{
    intto,cap,nxt;
} edge[EM];

int head[VM],gap[VM],dist[VM],cur[VM],pre[VM];
int ep,src,des,n;
void addedge (int cu,int cv,int cw)
{

    edge[ep].to= cv;
    edge[ep].cap= cw;
    edge[ep].nxt= head[cu];
    head[cu] =ep;
    ep ++;
    edge[ep].to= cu;
    edge[ep].cap= 0;
    edge[ep].nxt= head[cv];
    head[cv] =ep;
    ep ++;
}
int min (int a ,int b)
{
    return a> b ? b : a;
}

int sap ()
{

    memset(dist,0,sizeof(dist));
    memset(gap,0,sizeof (dist));
    memcpy(cur,head,sizeof(dist));
    int res =0;
    int u =pre[src] = src;
    int aug =inf;
    gap[0] =n;
    while(dist[src] < n)
    {
loop:
       for (int &i = cur[u]; i != -1; i =edge[i].nxt)
       {
           int v = edge[i].to;
           if (edge[i].cap && dist[u] ==dist[v] + 1)
           {
               aug = min (aug,edge[i].cap);
               pre[v] = u;
               u = v;
               if (v == des)
               {
                   res += aug;
                   for (u = pre[u]; v != src; v = u,u = pre[u])
                   {
                       edge[cur[u]].cap -= aug;
                       edge[cur[u]^1].cap += aug;
                   }
                   aug = inf; //
               }
               goto loop;
           }
       }
       int mindist = n;  //
       for (int i = head[u]; i != -1; i = edge[i].nxt)
       {
           int v = edge[i].to;
           if (edge[i].cap && mindist> dist[v])
           {
               cur[u] = i;
               mindist = dist[v];
           }
       }
       if ((--gap[dist[u]]) == 0)
           break;
       dist[u] = mindist + 1;
       gap[dist[u]] ++;
       u = pre[u];
    }
    returnres;
}

int main()
{
    int i,j;
    int N, F,D;
    while(~scanf("%d%d%d", &N, &F,&D))
    {
       ep = 0;
       memset(head, -1, sizeof(head));
       src = 0, des = 2*N + F + D +1;      //src: 0; food: 1--F; people: F+1--F+2*N;
       int f,d;                              //drink:F+2*N+1--F+2*N+D; des:F+2*N+D+1
       n = F + D + 2*N + 2;
       for (i=1; i!=F+1; ++i)
       {
           scanf("%d", &f);
           addedge(src, i, f);
       }
       for (i=F+2*N+1; i<=F+2*N+D; ++i)
       {
           scanf("%d", &d);
           addedge(i, des, d);
       }
       for (i=F+1; i<=F+N; ++i)
           addedge(i, i + N, 1);
       char op[205];
       for (i=1; i<=N; ++i)
       {
           scanf("%s", &op);
           for (j=1; j<=F; ++j)
           {
               if (op[j - 1] == 'Y')
                   addedge(j, F+i, 1);
           }
       }
       for (i=1; i<=N; ++i)
       {
           scanf("%s", &op);
           for(j=1; j<=D; ++j)
           {
               if (op[j - 1] == 'Y')
                   addedge(F+N+i, F+2*N+j, 1);
           }
       }
       printf("%d\n", sap());
    }
    return0;
}