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 = 2010;
const int M = 200010;
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 = 0;
    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 = 0;
        memset(head, -1, sizeof(head));
    }
    bool bfs(){
        queue<int> que;
        memset(level, -1, sizeof(level));
        level[S] = 0;
        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;
                int w = edge[i].w;
                if(level[v] == -1 && w > 0){
                    level[v] = level[u]+1;
                    que.push(v);
                }
            }
        }
        return level[T] != -1;
    }
    int dfs(int u, int flow){
        if(u == T)return flow;
        int ans = 0, fw;
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].to, w = edge[i].w;
            if(!w || level[v] != level[u]+1)
                  continue;
            fw = dfs(v, min(flow-ans, w));
            ans += fw;
            edge[i].w -= fw;
            edge[i^1].w += fw;
            if(ans == flow)return ans;
        }
        if(ans == 0)level[u] = -1;
        return ans;
    }
    int maxflow(){
        int flow = 0, f;
        while(bfs())
              while((f = dfs(S, INF)) > 0)
                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 = 0, t = 2*n+f+d+2;
        dinic.init(s, t);
        for(int i = 1; i <= n; i++)
              add_edge(i, n+i, 1);
        for(int i = 1; i <= f; i++){
            scanf("%d", &w);
              add_edge(s, 2*n+i, w);
        }
        for(int i = 1; i <= d; i++){
            scanf("%d", &w);
              add_edge(2*n+f+i, t, w);
        }
        for(int i = 1; i <= n; i++){
            scanf("%s", str);
            for(int j = 0; j < f; j++){
                if(str[j] == 'Y')
                    add_edge(2*n+j+1, i, 1);
            }
        }
        for(int i = 1; i <= n; i++){
            scanf("%s", str);
            for(int j = 0; j < d; j++){
                if(str[j] == 'Y')
                    add_edge(n+i, 2*n+f+j+1, 1);
            }
        }
        printf("%d
", dinic.maxflow());
    }    
    return 0;
}