hdu 4292 Food

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1054    Accepted Submission(s): 390

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

 

代码：

#include<cstdio>
#include<cstring>
const int N=1005;
const int M=200005;
const int INF=0x7fffffff;
int n,s,t,tol;
int head[N],arc[N],pre[N],dis[N],gap[N];
struct node
{
    int y,f,nxt;
}edge[M];
void add(int x,int y,int f)
{
    edge[tol].y=y;
    edge[tol].f=f;
    edge[tol].nxt=head[x];
    head[x]=tol++;

    edge[tol].y=x;
    edge[tol].f=0;
    edge[tol].nxt=head[y];
    head[y]=tol++;
}
void sap()
{
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    memcpy(arc,head,sizeof(dis));
    gap[0]=n;
    int ans=0,arg=INF,u=pre[s]=s;
    while(dis[s]<n)
    {
L:
        for(int &i=arc[u];i!=-1;i=edge[i].nxt)
        {
            int v=edge[i].y;
            if(edge[i].f&&dis[u]==dis[v]+1)
            {
                if(arg>edge[i].f)arg=edge[i].f;
                pre[v]=u;
                u=v;
                if(v==t)
                {
                    ans+=arg;
                    for(u=pre[u];v!=s;v=u,u=pre[u])
                    {
                        edge[arc[u]].f-=arg;
                        edge[arc[u]^1].f+=arg;
                    }
                    arg=INF;
                }
                goto L;
            }
        }
        int min=n;
        for(int i=head[u];i!=-1;i=edge[i].nxt)
        {
            int v=edge[i].y;
            if(edge[i].f&&min>dis[v])
            {
                arc[u]=i;
                min=dis[v];
            }
        }
        if(--gap[dis[u]]==0)break;
        dis[u]=min+1;
        gap[dis[u]]++;
        u=pre[u];
    }
    printf("%d\n",ans);
}
int main()
{
    int i,j,N,F,D;
    while(~scanf("%d%d%d",&N,&F,&D))
    {
        tol=0;
        memset(head,-1,sizeof(head));
        s=0;t=2*N+F+D+1;
        int f,d;
        n=F+D+2*N+2;
        for(i=1;i<=F;i++)
        {
            scanf("%d",&f);
            add(s,i,f);
        }
        for(i=F+2*N+1;i<=n-2;i++)
        {
            scanf("%d",&d);
            add(i,t,d);
        }
        for(i=F+1;i<=F+N;i++)add(i,i+N,1);
        char op[205];
        for(i=1;i<=N;i++)
        {
            scanf("%s",op+1);
            for(j=1;j<=F;j++)
            {
                if(op[j]=='Y')add(j,F+i,1);
            }
        }
        for(i=1;i<=N;i++)
        {
            scanf("%s",op+1);
            for(j=1;j<=D;j++)
            {
                if(op[j]=='Y')add(F+N+i,F+2*N+j,1);
            }
        }
        sap();
    }
    return 0;
}