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
题意：
有一些个数有限的 不同种类的糖果 和 一些不同种类的饮料 ， 每个人的口味不同，所以可以选择任意的糖果和饮料 ， 每个都只选一个
求能满足最多的人的数量
解析：
建立超级源点s和超级汇点t 把s和糖果连在一起，边权为糖果的数量， t和饮料连载一其，边权为饮料的数量， 然后把每一个人拆成两个点 其中边权为1
人和糖果、饮料 都建立边 边权为INF；
代码如下：
Dinic + 当前弧优化
#include <iostream> #include <cstring> #include <cstdio> #include <queue> #include <cmath> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; const int maxn =101000, INF = 0x7fffffff; int cnt = 0, s, t; int head[maxn], d[maxn], cur[maxn]; char str[300]; struct node{ int u, v, c, next; }Node[maxn*4]; void add_(int u, int v, int c) { Node[cnt].u = u; Node[cnt].v = v; Node[cnt].c = c; Node[cnt].next = head[u]; head[u] = cnt++; } void add(int u,int v,int c) { add_(u,v,c); add_(v,u,0); } bool bfs() { queue<int> Q; mem(d,0); Q.push(s); d[s] = 1; while(!Q.empty()) { int u = Q.front(); Q.pop(); for(int i=head[u]; i!=-1; i=Node[i].next) { node e = Node[i]; if(!d[e.v] && e.c > 0) { d[e.v] = d[e.u] + 1; // cout<< e.v << " " << d[e.v] <<endl; Q.push(e.v); if(e.v == t) return 1; } } } // cout<< d[t] <<endl; return d[t] != 0; } int dfs(int u,int cap) { if(u == t || cap == 0) return cap; int ret = 0; for(int &i=cur[u]; i!=-1; i=Node[i].next) { node e = Node[i]; if(d[e.v] == d[e.u] + 1 && e.c > 0) { int V = dfs(e.v, min(cap, e.c)); Node[i].c -= V; Node[i^1].c += V; cap -= V; ret += V; if(cap == 0) break; } } if(cap > 0) d[u] = -1; return ret; } int Dinic() { int ans = 0; while(bfs()) { memcpy(cur,head,sizeof(head)); ans += dfs(s,INF); } return ans; } int main() { int n, f, d; while(~scanf("%d%d%d",&n,&f,&d)) { cnt = 0; mem(head,-1); int temp; s = 0, t = f+d+n+n+10; for(int i=1; i<=f; i++) { scanf("%d",&temp); add(s,i,temp); } for(int i=1; i<=d; i++) { scanf("%d",&temp); add(f+i,t,temp); } for(int i=1; i<=n; i++) { scanf("%s",str); for(int j=0; j<f; j++) { if(str[j] == 'Y'){ add(j+1,f+d+i,INF); } } } for(int i=1; i<=n; i++) add(f+d+i, f+d+n+i,1); for(int i=1; i<=n; i++) { scanf("%s",str); for(int j=0; j<d; j++) { if(str[j] == 'Y') add(f+d+n+i,f+j+1,INF); } } cout<< Dinic() <<endl; } return 0; }