Food HDU - 4292 (结点容量 拆点) Dinic

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. 

Input　　There are several test cases. 
　　For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink. 
　　The second line contains F integers, the ith number of which denotes amount of representative food. 
　　The third line contains D integers, the ith number of which denotes amount of representative drink. 
　　Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no. 
　　Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no. 
　　Please process until EOF (End Of File). 
Output　　For each test case, please print a single line with one integer, the maximum number of people to be satisfied. 
Sample Input

4 3 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY

Sample Output

3

题意：

有一些个数有限的  不同种类的糖果 和 一些不同种类的饮料 ， 每个人的口味不同，所以可以选择任意的糖果和饮料 ， 每个都只选一个

求能满足最多的人的数量

解析：

建立超级源点s和超级汇点t  把s和糖果连在一起，边权为糖果的数量，  t和饮料连载一其，边权为饮料的数量， 然后把每一个人拆成两个点 其中边权为1

人和糖果、饮料 都建立边 边权为INF；

代码如下：

Dinic + 当前弧优化

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn =, INF = 0x7fffffff;
int cnt = , s, t;
int head[maxn], d[maxn], cur[maxn];
char str[];
struct node{
    int u, v, c, next;
}Node[maxn*];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u,int v,int c)
{
    add_(u,v,c);
    add_(v,u,);
}

bool bfs()
{
    queue<int> Q;
    mem(d,);
    Q.push(s);
    d[s] = ;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u]; i!=-; i=Node[i].next)
        {
            node e = Node[i];
            if(!d[e.v] && e.c > )
            {
                d[e.v] = d[e.u] + ;
           //     cout<< e.v << "   "  << d[e.v] <<endl;
                Q.push(e.v);
                if(e.v == t) return ;
            }
        }
    }
  //  cout<< d[t] <<endl;
    return d[t] != ;
}

int dfs(int u,int cap)
{
    if(u == t || cap == )
        return cap;
    int ret = ;
    for(int &i=cur[u]; i!=-; i=Node[i].next)
    {
        node e = Node[i];
        if(d[e.v] == d[e.u] +  && e.c > )
        {
            int V = dfs(e.v, min(cap, e.c));
            Node[i].c -= V;
            Node[i^].c += V;
            cap -= V;
            ret += V;
            if(cap == ) break;
        }
    }
    if(cap > ) d[u] = -;
    return ret;
}

int Dinic()
{
    int ans = ;
    while(bfs())
    {
        memcpy(cur,head,sizeof(head));
        ans += dfs(s,INF);
    }
    return ans;
}
int main()
{
    int n, f, d;
    while(~scanf("%d%d%d",&n,&f,&d))
    {
        cnt = ;
        mem(head,-);
        int temp;
        s = , t = f+d+n+n+;
        for(int i=; i<=f; i++)
        {
            scanf("%d",&temp);
            add(s,i,temp);
        }
        for(int i=; i<=d; i++)
        {
            scanf("%d",&temp);
            add(f+i,t,temp);
        }
        for(int i=; i<=n; i++)
        {
            scanf("%s",str);
            for(int j=; j<f; j++)
            {
                if(str[j] == 'Y'){
                    add(j+,f+d+i,INF);
                }
            }
        }
        for(int i=; i<=n; i++)
            add(f+d+i, f+d+n+i,);
        for(int i=; i<=n; i++)
        {
            scanf("%s",str);
            for(int j=; j<d; j++)
            {
                if(str[j] == 'Y')
                    add(f+d+n+i,f+j+,INF);
            }
        }
        cout<< Dinic() <<endl;
    }

    return ;
}