hdu 4292 最大流拆点建图

/*
题意：F种食物和D种饮料，每种食物和饮料的数目也是固定的，总共有N位顾客，每位顾客都只吃喝固定种类的食品
饮料，问最多能满足多少为顾客。

题解：最大流+拆点；
建图：将每位顾客拆成两个点，同一顾客之间加入权值为1的有向边限制了只能是一位一位顾客来满足，再加入源点
来连接每一种食物，权值为该食物数量，然后根据顾客喜欢的食物连接顾客，权值为1，因为每位顾客只需1份食物饮
料，然后加入饮料结点并根据顾客喜好连接，权值同样为1，然后将饮料与汇点连接，权值为饮料的数目，最终求得
的最大流即为最多能满足的顾客。
*/
#include <cstdio>
#include <cstring>

#define EMAX 200000
#define VMAX 1005

const int INF = 0xfffffff;

int head[VMAX],dis[VMAX],cur[VMAX],gap[VMAX],pre[VMAX];
int EN;
struct edge
{
    int from,to;
    int weight;
    int next;
}e[EMAX];

void insert(int u,int v,int w) 
{
    e[EN].to = v;
    e[EN].weight = w;
    e[EN].next = head[u];    
    head[u] = EN++;
    e[EN].weight = 0;
    e[EN].to = u;
    e[EN].next = head[v];    
    head[v] = EN++;
}

int sap(int s,int t, int n)
{
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    for(int i=0; i<=n; i++)
        cur[i] = head[i];
    int u = pre[s];
    pre[s] = s;
    int ret = 0;
    int temp = -1;
    gap[0] = n;
    bool flag;
    while(dis[s] < n)
    {
        flag = false;
        for(int &i = cur[u]; i != -1; i = e[i].next)
        {
            int v = e[i].to;
            if(e[i].weight && dis[u] == dis[v] + 1)
            {
                if (temp == -1 || temp>e[i].weight)
                    temp = e[i].weight;
                pre[v] = u;
                u = v;
                if(v == t)
                {
                    ret += temp;
                    for(u = pre[u];v != s;v = u,u = pre[u])
                    {
                        e[cur[u]].weight -= temp;
                        e[cur[u]^1].weight += temp;
                    }
                    temp = -1;
                }
                flag = true;
                break;
            }
        }
        if (flag)
            continue;

        int mindis = n;
        for(int i = head[u]; i != -1 ; i = e[i].next)
        {
            int v = e[i].to;
            if(e[i].weight && mindis > dis[v])
            {
                cur[u] = i;
                mindis = dis[v];
            }
        }
        gap[dis[u]]--;
        if( gap[dis[u]] == 0)
            break;
        dis[u] = mindis+1;
        gap[dis[u]]++;
        u = pre[u];
    }
    return ret;
}

int main(void)
{
    int n,f,d,t;
    char s[205];
    while (scanf("%d%d%d",&n,&f,&d) == 3)
    {
        memset(head,-1,sizeof(head));
        EN = 0;
        for(int i=1; i<=f; i++)
        {
            scanf("%d",&t);
            insert(0,i,t);//源点加边
        }
        for(int i=1; i<=d; i++)
        {
            scanf("%d",&t);
            insert(f+2*n+i,f+2*n+d+1,t);//汇点加边
        }
        for(int i=1; i<=n; i++)
            insert(f+i,f+n+i,1);//拆点
        for(int i=1; i<=n; i++)
        {
            scanf("%s",s+1);
            for(int j=1; j<=f; j++)
                if (s[j] == 'Y')
                    insert(j,f+i,1);//顾客与食物加边
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%s",s+1);
            for(int j=1; j<=d; j++)
                if (s[j] == 'Y')
                    insert(f+n+i,f+2*n+j,1);//顾客与饮料加边
        }
        printf("%d\n",sap(0,f+2*n+d+1,f+2*n+d+2));
    }
    return 0;
}