hdu-4292 Food(最大流）dinic

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6598    Accepted Submission(s): 2238
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

题意：一共有n个人f种食物，d种饮料，每个人对每种饮料和食物喜欢就是‘Y’,不喜欢是‘N’,求最多有几个人会满足。

思路：最大流，拆人。

这个题比较容易超时，好像一般都用了邻接表才能过。

下面代码G++会超时，C++可以过。

#include<map>
#include<set>
#include<queue>
#include<math.h>
#include <cstdio>
#include <ctime>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f3f
#define ll long long
#define OPT(_)(hip+(((_)-hip)^1))
#define maxn 300000
using namespace std;
struct node  {
    int v,w;
    struct node *next;
}hip[maxn],*hop;
node *h[maxn];
int d[maxn];
int s,t;
void w(int u,int v,int w,int b){
  hop->v=v;
  hop->w=w;
  hop->next=h[u];
  h[u]=hop++;
  hop->v=u;
  hop->w=b;
  hop->next=h[v];
  h[v]=hop++;
}
bool bfs(){
    queue<int>q;
    while(!q.empty())q.pop();
    memset(d,0,sizeof(d));
    d[s]=1;
    q.push(s);
    do{
        int u=q.front();
        q.pop();
        struct node*T;
        for(T=h[u];T!=NULL;T=T->next){
            if((T->w)>0&&d[T->v]==0){
                d[T->v]=d[u]+1;
                q.push(T->v);
            }
        }
    }while(!q.empty());
    if(d[t]==0)return false;
    else
        return true;
}
int  dfs(int u,int dist){
    if(u==t)
        return dist;
    struct node *p;
    for(p=h[u];p!=NULL;p=p->next){
        if((d[p->v]==d[u]+1)&&(p->w)>0){
            int di=dfs(p->v,min(dist,p->w));
            if(di>0){
             p->w-=di;
             OPT(p)->w+=di;
             return di;
            }
        }
    }
    return 0;
}
int dinic(){
    int ans=0;
    while(bfs()){
        while(int d=dfs(s,inf)){
            ans+=d;}
    }
    return ans;}
int main(){
    int n,f,d;
    while(~scanf("%d%d%d",&n,&f,&d)){
    s=0;t=n+n+d+f+1;
    memset(hip,0,sizeof(hip));
    hop=hip;
    for(int i=0;i<maxn;i++){h[i]=NULL;}
    for(int i=1;i<=f;i++){
        int x;scanf("%d",&x);w(s,i,x,0);
    }
    for(int i=1;i<=n;i++){
        w(f+i,f+n+i,1,0);
    }
    for(int i=1;i<=d;i++){
        int x;scanf("%d",&x);w(f+n+n+i,t,x,0);
    }
    for(int i=1;i<=n;i++){
        string s;cin>>s;
    for(int j=1;j<=f;j++){
        if(s[j-1]=='Y'){w(j,f+i,1,0);}}
    }
     for(int i=1;i<=n;i++){
        string s;cin>>s;
    for(int j=1;j<=d;j++){
        if(s[j-1]=='Y'){w(f+i+n,f+n+n+j,1,0);}}
    }
    cout<<dinic()<<endl;
  }
}