Hdu 4292 Food (maximum Stream)

Hdu 4292 Food (maximum Stream)
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 your 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 cocould 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 your people as possible. so you must decide a plan where some people are served while requirements of the rest of them are unmet. you shoshould notice that, when one's requirement is unmet, he/she woshould 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 wocould 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 wowould 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

The number of guests, the type of food, the type of drinks, the stock of each food and beverage, and the requirements of each customer are given. Find the maximum number of guests to meet. Solution: poj 3281 Dining is a type of question. Set the super source point, connect to all the food, the capacity is the inventory of the food, set the super collection point, so that all drinks point to the super collection point, the capacity is the inventory of the drink. Then the customer connects the corresponding food and drinks with a capacity of 1.

#include 
  
   #include 
   
    #include #include 
    
     #include 
     
      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, dinic());        }    return 0;}