HDU4292(KB11-H 最大流)

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5945    Accepted Submission(s): 2010

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

 

Source

2012 ACM/ICPC Asia Regional Chengdu Online

 

建图与POJ3281一毛一样，边开少了狂T。。。

 //2017-08-24
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <queue>
 #pragma comment(linker, "/STACK:1024000000,1024000000") 

 using namespace std;

 const int N = ;
 const int M = ;
 const int INF = 0x3f3f3f3f;
 int head[N], tot;
 struct Edge{
     int next, to, w;
 }edge[M];

 void add_edge(int u, int v, int w){
     edge[tot].w = w;
     edge[tot].to = v;
     edge[tot].next = head[u];
     head[u] = tot++;

     edge[tot].w = ;
     edge[tot].to = u;
     edge[tot].next = head[v];
     head[v] = tot++;
 }

 struct Dinic{
     int level[N], S, T;
     void init(int _S, int _T){
         S = _S;
         T = _T;
         tot = ;
         memset(head, -, sizeof(head));
     }
     bool bfs(){
         queue<int> que;
         memset(level, -, sizeof(level));
         level[S] = ;
         que.push(S);
         while(!que.empty()){
             int u = que.front();
             que.pop();
             for(int i = head[u]; i != -; i = edge[i].next){
                 int v = edge[i].to;
                 int w = edge[i].w;
                 if(level[v] == - && w > ){
                     level[v] = level[u]+;
                     que.push(v);
                 }
             }
         }
         return level[T] != -;
     }
     int dfs(int u, int flow){
         if(u == T)return flow;
         int ans = , fw;
         for(int i = head[u]; i != -; i = edge[i].next){
             int v = edge[i].to, w = edge[i].w;
             if(!w || level[v] != level[u]+)
                   continue;
             fw = dfs(v, min(flow-ans, w));
             ans += fw;
             edge[i].w -= fw;
             edge[i^].w += fw;
             if(ans == flow)return ans;
         }
         if(ans == )level[u] = -;
         return ans;
     }
     int maxflow(){
         int flow = , f;
         while(bfs())
               while((f = dfs(S, INF)) > )
                 flow += f;
         return flow;
     }
 }dinic;

 char str[N];

 int main()
 {
     //std::ios::sync_with_stdio(false);
     //freopen("inputH.txt", "r", stdin);
     int n, f, d, w;
     while(scanf("%d%d%d", &n, &f, &d) != EOF){
         int s = , t = *n+f+d+;
         dinic.init(s, t);
         for(int i = ; i <= n; i++)
               add_edge(i, n+i, );
         for(int i = ; i <= f; i++){
             scanf("%d", &w);
               add_edge(s, *n+i, w);
         }
         for(int i = ; i <= d; i++){
             scanf("%d", &w);
               add_edge(*n+f+i, t, w);
         }
         for(int i = ; i <= n; i++){
             scanf("%s", str);
             for(int j = ; j < f; j++){
                 if(str[j] == 'Y')
                     add_edge(*n+j+, i, );
             }
         }
         for(int i = ; i <= n; i++){
             scanf("%s", str);
             for(int j = ; j < d; j++){
                 if(str[j] == 'Y')
                     add_edge(n+i, *n+f+j+, );
             }
         }
         printf("%d\n", dinic.maxflow());
     }
     return ;
 }