hdu 4292 Food （最大流）

hdu 4292 Food

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：给出客人数量，食物种类，饮品种类，每种食物和饮品的库存，还有每位顾客的要求。求出，最多能满足多少位客人的要求。

解题思路：跟poj 3281 Dining是一种类型的题目。 设置超级源点，连向所有的食物，容量为该食物的库存，设置超级汇点，使所有饮品指向超级汇点，容量为该饮品的库存。然后客户连接对应的食物和饮品，容量为1。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;

const int N = 20005;
const int M = 200005;
const int OF = 300;
const int INF = 0x3f3f3f3f;
const int FIN = 1999;
typedef long long ll;
int n, f, d, s, t;
int ec, head[N], first[N], que[N], lev[N];
int Next[M], to[M], v[M];

void addEdge(int a,int b,int c) {
    to[ec] = b;
    v[ec] = c;
    Next[ec] = first[a];
    first[a] = ec++;

    to[ec] = a;
    v[ec] = 0;
    Next[ec] = first[b];
    first[b] = ec++;
}

int BFS() {
    int kid, now, f = 0, r = 1, i;
    memset(lev, 0, sizeof(lev));
    que[0] = s, lev[s] = 1;
    while (f < r) {
        now = que[f++];
        for (i = first[now]; i != -1; i = Next[i]) {
            kid = to[i];    
            if (!lev[kid] && v[i]) {
                lev[kid] = lev[now] + 1;    
                if (kid == t) return 1;
                que[r++] = kid;
            }
        }
    }
    return 0;
}

int DFS(int now, int sum) {
    int kid, flow, rt = 0;
    if (now == t) return sum;
    for (int i = head[now]; i != -1 && rt < sum; i = Next[i]) {
        head[now] = i;  
        kid = to[i];
        if (lev[kid] == lev[now] + 1 && v[i]) {
            flow = DFS(kid, min(sum - rt, v[i]));
            if (flow) {
                v[i] -= flow;
                v[i^1] += flow;
                rt += flow;
            } else lev[kid] = -1;   
        }           
    }
    return rt;
}

int dinic() {
    int ans = 0;
    while (BFS()) {
        for (int i = 0; i <= t; i++) {
            head[i] = first[i];
        }           
        ans += DFS(s, INF);
    }
    return ans;
}   

char c[205];
void input() {
    int a;
    for (int i = 1; i <= f; i++) {
        scanf("%d", &a);
        addEdge(s, i, a);
    }
    for (int i = 1; i <= d; i++) {
        scanf("%d", &a);
        addEdge(i + 3 * OF, t, a);
    }
    for (int i = 1; i <= n; i++) {
        addEdge(i + OF, i + 2 * OF, 1);
    }       
    getchar();
    for (int i = 1; i <= n; i++) {
        scanf("%s", c);
        for (int j = 0; j < f; j++) {
            if (c[j] == ‘Y‘) {
                addEdge(j + 1, i + OF, 1);
            }       
        }   
    }
    for (int i = 1; i <= n; i++) {
        scanf("%s", c); 
        for (int j = 0; j < d; j++) {
            if (c[j] == ‘Y‘) {
                addEdge(i + 2 * OF, (j + 1) + 3 * OF, 1);   
            }
        }   
    }
}

int main() {
    while (scanf("%d %d %d", &n, &f, &d) == 3) {
        ec = 0;
        memset(first, -1, sizeof(first));
        s = 0, t = FIN;
        input();
        printf("%d\n", dinic());    
    }
    return 0;
}