hdu 4292 Food 最大流

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4292

You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
　　The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
　　You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
　　Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.

题意描述：你准备了F种食物（F<=200）和D种饮料（D<=200），每种食物和饮料都有一定的数量。有n个人（n<=200），每个人都有喜欢的食物和饮料并且只能吃自己喜欢的食物、喝喜欢的饮料，求出能够吃到喜欢的食物并喝到喜欢的饮料的最大人数。

算法分析：运用网络最大流求解即可。新增源点from和汇点to，from->食物（w为食物数量），饮料->to（w为饮料数量），对每个人 i 拆边 i 和 i+n ，喜欢的食物->i（w为1），i -> i+n（w为1），i+n->喜欢的饮料（w为1）。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#define inf 0x7fffffff
using namespace std;
const int maxn=200+10;
const int M = 999999;

int n,F,D;
int from,to,d[maxn*10];
struct node
{
    int v,flow;
    int next;
}edge[M*2];
int head[maxn*10],edgenum;

void add(int u,int v,int flow)
{
    edge[edgenum].v=v ;edge[edgenum].flow=flow;
    edge[edgenum].next=head[u];
    head[u]=edgenum++;

    edge[edgenum].v=u ;edge[edgenum].flow=0;
    edge[edgenum].next=head[v];
    head[v]=edgenum++;
}

int bfs()
{
    memset(d,0,sizeof(d));
    d[from]=1;
    queue<int> Q;
    Q.push(from);
    while (!Q.empty())
    {
        int u=Q.front() ;Q.pop() ;
        for (int i=head[u] ;i!=-1 ;i=edge[i].next)
        {
            int v=edge[i].v;
            if (!d[v] && edge[i].flow>0)
            {
                d[v]=d[u]+1;
                Q.push(v);
                if (v==to) return 1;
            }
        }
    }
    return 0;
}

int dfs(int u,int flow)
{
    if (u==to || flow==0) return flow;
    int cap=flow;
    for (int i=head[u] ;i!=-1 ;i=edge[i].next)
    {
        int v=edge[i].v;
        if (d[v]==d[u]+1 && edge[i].flow>0)
        {
            int x=dfs(v,min(cap,edge[i].flow));
            cap -= x;
            edge[i].flow -= x;
            edge[i^1].flow += x;
            if (cap==0) return flow;
        }
    }
    return flow-cap;
}

int dinic()
{
    int sum=0;
    while (bfs()) sum += dfs(from,inf);
    return sum;
}

int main()
{
    while (scanf("%d%d%d",&n,&F,&D)!=EOF)
    {
        memset(head,-1,sizeof(head));
        edgenum=0;
        from=F+2*n+D+1;
        to=from+1;
        int a;
        for (int i=1 ;i<=F ;i++)
        {
            scanf("%d",&a);
            add(from,i,a);
        }
        for (int i=1 ;i<=D ;i++)
        {
            scanf("%d",&a);
            add(F+2*n+i,to,a);
        }
        char str[maxn];
        memset(str,0,sizeof(str));
        for (int i=1 ;i<=n ;i++)
        {
            scanf("%s",str+1);
            for (int j=1 ;j<=F ;j++)
            {
                if (str[j]=='Y')
                    add(j,F+i,1);
            }
        }
        for (int i=1 ;i<=n ;i++)
        {
            scanf("%s",str+1);
            for (int j=1 ;j<=D ;j++)
            {
                if (str[j]=='Y')
                    add(F+n+i,F+2*n+j,1);
            }
        }
        for (int i=1 ;i<=n ;i++)
            add(F+i,F+n+i,1);
        printf("%d
",dinic());
    }
    return 0;
}