hdu 4292 网络最大流

http://acm.hdu.edu.cn/showproblem.php?pid=4292

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 31 1 11 1 1YYNNYYYNYYNYYNYYYNYYNNNY

 

Sample Output

3

/**hdu 4292 网络最大流题目大意：有F种食物，D种饮料，各有一定数量n个人，现在要求每个人的要求是1份食物1个饮料，问最多能满足多少人的要求解题思路：一看就是网络流的题。         建边思想：把n个人对应拆成个连个点i和i‘，之间连边权值1；                   每种食物和喜欢其的i之间权值为1的边                   每个人和他喜欢的饮料之间连权值为1的边                   起点和每种食物之间连权值为该种食物数量的边；                   每种饮料喝终点之间连权值为该种饮料数量的边。         跑一遍最大流即可。*/#include<cstdio>#include<iostream>#include<algorithm>#include<string.h>using namespace std;const int oo=1e9;const int mm=8811111;const int mn=1212;int node,src,dest,edge;int ver[mm],flow[mm],_next[mm];int head[mn],work[mn],dis[mn],q[mn];void prepare(int _node,int _src,int _dest){    node=_node,src=_src,dest=_dest;    for(int i=0; i<node; ++i)head[i]=-1;    edge=0;}void addedge(int u,int v,int c){    ver[edge]=v,flow[edge]=c,_next[edge]=head[u],head[u]=edge++;    ver[edge]=u,flow[edge]=0,_next[edge]=head[v],head[v]=edge++;}bool Dinic_bfs(){    int i,u,v,l,r=0;    for(i=0; i<node; ++i)dis[i]=-1;    dis[q[r++]=src]=0;    for(l=0; l<r; ++l)        for(i=head[u=q[l]]; i>=0; i=_next[i])            if(flow[i]&&dis[v=ver[i]]<0)            {                dis[q[r++]=v]=dis[u]+1;                if(v==dest)return 1;            }    return 0;}int Dinic_dfs(int u,int exp){    if(u==dest)return exp;    for(int &i=work[u],v,tmp; i>=0; i=_next[i])        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)        {            flow[i]-=tmp;            flow[i^1]+=tmp;            return tmp;        }    return 0;}int Dinic_flow(){    int i,ret=0,delta;    while(Dinic_bfs())    {        for(i=0; i<node; ++i)work[i]=head[i];        while(delta=Dinic_dfs(src,oo))ret+=delta;    }    return ret;}int n,D,F,f[205],d[205];char a[1005];int main(){    while(~scanf("%d%d%d",&n,&F,&D))    {        prepare(n*2+F+D+2,0,F+D+2*n+1);        for(int i=1;i<=F;i++)        {            int x;            scanf("%d",&x);            addedge(src,i,x);        }        for(int i=1;i<=D;i++)        {            int x;            scanf("%d",&x);            addedge(i+F+n*2,dest,x);        }        for(int i=1;i<=n;i++)        {            addedge(i+F,i+n+F,1);            scanf("%s",a+1);            for(int j=1;j<=F;j++)            {                if(a[j]=='Y')                    addedge(j,i+F,1);            }        }        for(int i=1;i<=n;i++)        {            scanf("%s",a+1);            for(int j=1;j<=D;j++)            {                if(a[j]=='Y')                    addedge(i+F+n,F+n*2+j,1);            }        }        printf("%d\n",Dinic_flow());    }    return 0;}