hdu 4292 Food 【图论-网络流-最大流-Dinic】

                                    Food    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description
　　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

题目大意：有n个人，每个人都有自己喜欢的食物和饮料，先有F中食物，D中饮料，并给出每个人的喜好，让你求所给食物和饮料最多能满足多少人的需求
注：本题和poj3281差不多，现给出我做poj3281时的文章链接：
http://blog.csdn.net/xingdragon/article/details/70237266
解题思路也在上面链接的文章里

知识点：最大流

**所用算法：**Dinic

AC代码：

//Dinic             AC//Ford-Fulkerson    Time limit//EK                Time limit# include <cstdio># include <cstring># include <queue>using namespace std;# define MAXN 10005# define MAXM 200005  //数组开小了 超时了好多次# define INF 1 << 30struct EDGE{    int to;    int w;    int next;}edge[MAXM];int tot;int head[MAXN];int dis[MAXN];int min(int a, int b){    return a > b ? b : a;}void Init(){    tot = 0;    memset(head, -1, sizeof(head));}void Addedge(int u, int v, int w){    edge[tot].to = v;    edge[tot].w = w;    edge[tot].next = head[u];    head[u] = tot++;    edge[tot].to = u;    edge[tot].w = 0;    edge[tot].next = head[v];    head[v] = tot++;}bool Bfs(int s, int t){    memset(dis, 0, sizeof(dis));    queue<int> que;    dis[s] = 1;    que.push(s);    while (!que.empty())    {        int u = que.front(); que.pop();        for (int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].to;            if (dis[v] == 0 && edge[i].w > 0)            {                dis[v] = dis[u] + 1;                if (v == t)                {                    return true;                }                que.push(v);            }        }    }    return false;}int Dfs(int u, int t, int f){    if (u == t)    {        return f;    }    int cost = 0;    for (int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].to;        int w = edge[i].w;        if (dis[v] == dis[u] + 1 && w > 0)        {            int d = Dfs(v, t, min(w, f - cost));            if (d > 0)            {                edge[i].w -= d;                edge[i ^ 1].w += d;                cost += d;                if (cost == f)                {                    break;                }                else                 {                    dis[v] = -1;                }            }        }    }    return cost;}int Dinic(int s, int t){    int maxflow = 0;    while (Bfs(s, t))    {        maxflow += Dfs(s, t, INF);    }    return maxflow;}int main(void){    int n, f, d;    while (scanf("%d %d %d", &n, &f, &d) != EOF)    {        Init();        int i, j;        int s = 0;        int t = n * 2 + f + d + 1;        for (i = 1; i <= f; i++)        {            int w;            scanf("%d", &w);            Addedge(s, i, w);        }        for (i = 1; i <= d; i++)        {            int w;            scanf("%d", &w);            Addedge(f + 2 * n + i, t, w);        }        for (i = 1; i <= n; i++)        {            Addedge(f + i, f + n + i, 1);        }        char str[MAXN];        for (i = 1; i <= n; i++)        {            scanf("%s", &str[1]);            for (j = 1; j <= f; j++)            {                if (str[j] == 'Y')                {                    Addedge(j, f + i, 1);                }            }        }        for (i = 1; i <= n; i++)        {            scanf("%s", &str[1]);            for (j = 1; j <= d; j++)            {                if (str[j] == 'Y')                {                    Addedge(f + n + i, f + 2 * n + j, 1);                }            }        }        printf("%d\n", Dinic(s, t));    }    return 0;}