HDU

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

题解：

最大流送分题。不多说了希望今年网络赛也能出这样的题。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 910;

struct Edge{
	int flow,to,rev;
	Edge(){}
	Edge(int a,int b,int c):to(a),flow(b),rev(c){}
};

vector<Edge> E[MAXN];

inline void Add(int from,int to,int flow){
	E[from].push_back(Edge(to,flow,E[to].size()));
	E[to].push_back(Edge(from,0,E[from].size()-1));
}

int deep[MAXN];
int iter[MAXN];

bool BFS(int from,int to){
	memset(deep,-1,sizeof deep);
	deep[from] = 0;
	queue<int> Q;
	Q.push(from);
	while(!Q.empty()){
		int t = Q.front();
		Q.pop();
		for(int i=0 ; i<E[t].size() ; ++i){
			Edge& e = E[t][i];
			if(e.flow > 0 && deep[e.to] == -1){
				deep[e.to] = deep[t] + 1;
				Q.push(e.to);
			}
		}
	}
	return deep[to] != -1;
}

int DFS(int from,int to,int flow){
	if(from == to || flow == 0)return flow;
	for(int& i=iter[from] ; i<E[from].size() ; ++i){
		Edge& e = E[from][i];
		if(e.flow > 0 && deep[e.to] == deep[from] + 1){
			int nowflow = DFS(e.to,to,min(flow,e.flow));
			if(nowflow > 0){
				e.flow -= nowflow;
				E[e.to][e.rev].flow += nowflow;
				return nowflow;
			}
		}
	}
	return 0;
}

int Dinic(int from,int to){
	int sumflow = 0;
	while(BFS(from,to)){
		memset(iter,0,sizeof iter);
		int mid;
		while((mid = DFS(from,to,INF)) > 0)sumflow += mid;
	}
	return sumflow;
}

int main(){
	
	int N,F,D;
	char s[205];
	while(scanf("%d %d %d",&N,&F,&D)!=EOF){
		for(int i=1 ; i<=N ; ++i)Add(i,i+200,1);
		for(int i=1 ; i<=F ; ++i){
			int t;
			scanf("%d",&t);
			Add(0,i+400,t);
		}
		for(int i=1 ; i<=D ; ++i){
			int t;
			scanf("%d",&t);
			Add(i+600,MAXN-1,t);
		}
		for(int i=1 ; i<=N ; ++i){
			scanf("%s",s);
			for(int j=0 ; j<F ; ++j){
				if(s[j] == 'Y')Add(j+1+400,i,INF);
			}
		}
		for(int i=1 ; i<=N ; ++i){
			scanf("%s",s);
			for(int j=0 ; j<D ; ++j){
				if(s[j] == 'Y')Add(i+200,j+1+600,INF);
			}
		}
		printf("%d
",Dinic(0,MAXN-1));
		for(int i=0 ; i<MAXN ; ++i)E[i].clear();
	}
	
	return 0;
}