hdu4292

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1668    Accepted Submission(s): 606

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

 

Recommend

liuyiding

 

每个满足的人只要提供一份食物，和一份饮料即可，

将人拆点，超级源点，连食物，边权为食物数量，超级汇点连饮料，边权为数量，将食物与人一点相连，另一点与饮料相连

求最大流

#include <cstring>
#include <cstdio>
#include <queue>
#define MAXN 50000
#define MAXM 500000
#define inf 0x3f3f3f3f
using namespace std;
struct node
{
    int u,v,f;
};
node e[MAXM*3];
int first[MAXN],next[MAXM*3];
int gap[MAXN],d[MAXN],curedge[MAXN],pre[MAXN];
int cc;
inline void add_edge(int u,int v,int f)
{
    e[cc].u=u;
    e[cc].v=v;
    e[cc].f=f;
    next[cc]=first[u];
    first[u]=cc;
    cc++;

    e[cc].u=v;
    e[cc].v=u;
    e[cc].f=0;
    next[cc]=first[v];
    first[v]=cc;
    cc++;

}
int ISAP(int s,int t,int n)
{
    int cur_flow,flow_ans=0,u,tmp,neck,i,v;
    memset(d,0,sizeof(d));
    memset(gap,0,sizeof(gap));
    memset(pre,-1,sizeof(pre));
    for(i=0;i<=n;i++)
        curedge[i]=first[i];
    gap[0]=n+1;
    u=s;
    while(d[s]<=n)
    {
        if(u==t)
        {
            cur_flow=inf;
            for(i=s;i!=t;i=e[curedge[i]].v)
            {
                if(cur_flow>e[curedge[i]].f)
                {
                    neck=i;
                    cur_flow=e[curedge[i]].f;
                }
            }
            for(i=s;i!=t;i=e[curedge[i]].v)
            {
                tmp=curedge[i];
                e[tmp].f-=cur_flow;
                e[tmp^1].f+=cur_flow;
            }
            flow_ans+=cur_flow;
            u=neck;
        }
        for(i=curedge[u];i!=-1;i=next[i])
        {
            v=e[i].v;
            if(e[i].f&&d[u]==d[v]+1)
                break;
        }
        if(i!=-1)
        {
            curedge[u]=i;
            pre[v]=u;
            u=v;
        }
        else
        {
            if(0==--gap[d[u]])
                break;
            curedge[u]=first[u];
            for(tmp=n+5,i=first[u];i!=-1;i=next[i])
                if(e[i].f)
                    tmp=min(tmp,d[e[i].v]);
            d[u]=tmp+1;
            ++gap[d[u]];
            if(u!=s)
                u=pre[u];
        }
    }
    return flow_ans;
}
int main()
{
    int N,F,D;

    while(scanf("%d%d%d",&N,&F,&D)!=EOF)
    {
        memset(first,-1,sizeof(first));
        memset(next,-1,sizeof(next));
        cc=0;
        int s=0,t=F+D+N+N+1;
        int i;
        for(i=1;i<=F;i++)
        {
            int a;
            scanf("%d",&a);
            add_edge(s,i,a);
        }
        for(i=1;i<=D;i++)
        {
            int a;
            scanf("%d",&a);
            add_edge(i+F,t,a);
        }
        for(i=1;i<=N;i++)
        {
            add_edge(F+D+i,F+D+i+N,1);
        }
        char str[250];
        for(i=1;i<=N;i++)
        {
            scanf("%s",str+1);
            int j;
            for(j=1;j<=F;j++)
            {
                if(str[j]=='Y')
                {
                    add_edge(j,F+D+i,1);
                }
            }
        }
        for(i=1;i<=N;i++)
        {
            scanf("%s",str+1);
            int j;
            for(j=1;j<=D;j++)
            {
                if(str[j]=='Y')
                {
                    add_edge(F+D+i+N,F+j,1);
                }
            }
        }
        int res=ISAP(s,t,t);
        printf("%d\n",res);
    }

    return 0;
}