HDU 4292 Food

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2136    Accepted Submission(s): 745

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

构图是老生常谈的事情了。

S-->food连边food[i] ;

food-->man1连边1

man1-->man2连边1

mam2-->drink 连边1 ，

diink-->T连边dink[i]

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <stack>
#include <math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
#define MAXN 1000
#define inf 2100000000
int g[MAXN][MAXN],level[MAXN];
int bfs(int n,int s,int t){
    int que[1000],head,tail,i;
    head=tail=0;
    que[tail++]=s;
    memset(level,-1,sizeof(level));
    level[s]=1;
    while(head<tail){
        int x=que[head++];
        for(i=1;i<=n;++i)
            if((g[x][i])&&(level[i]==-1)){
                level[i]=level[x]+1;
                que[tail++]=i;
            }
    }
    return level[t]+1;
}
/*n为最大点，s起点 ，t终点 ,图中没有定点0，从1开始到n*/
int dinic(int n,int s,int t){
    int maxflow=0,i;
    while(bfs(n,s,t)){
        int pos,path[MAXN],cut=0,low,find;
        pos=0;
        path[pos++]=s;find=1;
        while(1){
            find=1;
            low=inf;
            while((path[pos-1]!=t)&&(find)){
                find=0;
                for(i=1;i<=n;++i)
                    if((level[i]==level[path[pos-1]]+1)&&(g[path[pos-1]][i])){
                        path[pos++]=i;
                        find=1;
                        break;
                    }
                    else if (i==n) break;
            }
            if(path[pos-1]==t){
                for(i=0;i<pos-1;++i){
                    if(low>g[path[i]][path[i+1]]){
                          low=g[path[i]][path[i+1]];
                          cut=i;
                    }
                }
                maxflow+=low;
                for(i=0;i<pos-1;++i){
                    g[path[i]][path[i+1]]-=low;
                    g[path[i+1]][path[i]]+=low;
                }
                pos=cut+1;
            }
            else{
                if(pos==1)
                    break;
                else{
                    level[path[pos-1]]=-1;
                    pos--;
                }
            }
        }
    }
    return maxflow;
}

const int Max_N=208 ;
char str[Max_N][Max_N] ;
int food[Max_N] ;
int drink[Max_N] ;
int S ,T  ;
int main(){
   int F , N , D ;
   while(scanf("%d%d%d",&N,&F,&D)!=EOF){
        memset(g,0,sizeof(g)) ;
        for(int i=1;i<=F;i++)
            scanf("%d",&food[i]) ;
        for(int i=1; i<=D ;i++)
            scanf("%d",&drink[i]) ;
        for(int i=1;i<=N;i++)
            scanf("%s",str[i]+1) ;
        for(int i=1;i<=N;i++)
           for(int j=1;j<=F;j++){
             if(str[i][j]=='Y')
                 g[j][F+i]=1 ;
        }
        for(int i=1;i<=N;i++)
            scanf("%s",str[i]+1) ;
        for(int i=1;i<=N;i++)
           for(int j=1;j<=D;j++){
             if(str[i][j]=='Y')
                 g[F+N+i][F+N+N+j]=1 ;
        }
        for(int i=1;i<=N;i++)
            g[F+i][F+N+i]=1 ;
        T=F+N+N+D+1 ;
        S=F+N+N+D+1+1 ;
        for(int i=1;i<=D;i++)
            g[F+N+N+i][T]=drink[i] ;
        for(int i=1;i<=F;i++)
            g[S][i]=food[i] ;
        cout<<dinic(S,S,T)<<endl ;
   }
   return 0 ;
}