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. 

4 3 3

1 1 1

1 1 1

YYN

NYY

YNY

YNY

YNY

YYN

YYN

NNY

3

题目大意：
这个就是一个牛吃草问题

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#include <iostream>

#include <queue>

#include <vector>

#include <map>

#include <cstring>

#include <string>

#define inf 0x3f3f3f3f

using namespace std;

const int maxn = 1e5 + ;

const int INF = 0x3f3f3f3f;

struct edge

{

    int u, v, c, f;

    edge(int u, int v, int c, int f) :u(u), v(v), c(c), f(f) {}

};

vector<edge>e;

vector<int>G[maxn];

int level[maxn];//BFS分层，表示每个点的层数

int iter[maxn];//当前弧优化

int m, s, t;

void init(int n)

{

    for (int i = ; i <= n; i++)G[i].clear();

    e.clear();

}

void add(int u, int v, int c)

{

    e.push_back(edge(u, v, c, ));

    e.push_back(edge(v, u, , ));

    m = e.size();

    G[u].push_back(m - );

    G[v].push_back(m - );

}

void BFS(int s)//预处理出level数组

//直接BFS到每个点

{

    memset(level, -, sizeof(level));

    queue<int>q;

    level[s] = ;

    q.push(s);

    while (!q.empty())

    {

        int u = q.front();

        q.pop();

        for (int v = ; v < G[u].size(); v++)

        {

            edge& now = e[G[u][v]];

            if (now.c > now.f && level[now.v] < )

            {

                level[now.v] = level[u] + ;

                q.push(now.v);

            }

        }

    }

}

int dfs(int u, int t, int f)//DFS寻找增广路

{

    if (u == t)return f;//已经到达源点，返回流量f

    for (int &v = iter[u]; v < G[u].size(); v++)

        //这里用iter数组表示每个点目前的弧，这是为了防止在一次寻找增广路的时候，对一些边多次遍历

        //在每次找增广路的时候，数组要清空

    {

        edge &now = e[G[u][v]];

        if (now.c - now.f >  && level[u] < level[now.v])

            //now.c - now.f > 0表示这条路还未满

            //level[u] < level[now.v]表示这条路是最短路，一定到达下一层，这就是Dinic算法的思想

        {

            int d = dfs(now.v, t, min(f, now.c - now.f));

            if (d > )

            {

                now.f += d;//正向边流量加d

                e[G[u][v] ^ ].f -= d;

                //反向边减d，此处在存储边的时候两条反向边可以通过^操作直接找到

                return d;

            }

        }

    }

    return ;

}

int Maxflow(int s, int t)

{

    int flow = ;

    for (;;)

    {

        BFS(s);

        if (level[t] < )return flow;//残余网络中到达不了t，增广路不存在

        memset(iter, , sizeof(iter));//清空当前弧数组

        int f;//记录增广路的可增加的流量

        while ((f = dfs(s, t, INF)) > )

        {

            flow += f;

        }

    }

    return flow;

}

int main()

{

    int n, fo, di;

    while(scanf("%d%d%d", &n, &fo, &di)!=EOF)

    {

        init(*n+fo+di+);

        s = , t =  * n + fo + di + ;

        for (int i = ; i <= fo; i++)

        {

            int num;

            scanf("%d", &num);

            add(s, i, num);

        }

        for (int i = ; i <= di; i++)

        {

            int num;

            scanf("%d", &num);

            add(i + fo +  * n, t, num);

        }

        for (int i = ; i <= n; i++)

        {

            add(i + fo, i + n + fo, );

        }

        for (int i = ; i <= n; i++)

        {

            char qw[];

            scanf("%s", qw);

            for (int j = ; j < fo; j++)

            {

                if (qw[j] == 'Y') add(j + , i + fo, );

            }

        }

        for (int i = ; i <= n; i++)

        {

            char qw[];

            scanf("%s", qw);

            for (int j = ; j < di; j++)

            {

                if (qw[j] == 'Y') add(i + fo + n, j + fo +  * n + , );

            }

        }

        int ans = Maxflow(s, t);

        printf("%d\n", ans);

    }

    return ;

}