POJ - 3281 Dining（拆点+最大网络流）

Dining

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 18230		Accepted: 8132

Description

Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.

Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.

Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.

Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).

Input

Line 1: Three space-separated integers: N, F, and D
Lines 2..N+1: Each line i starts with a two integers F_i and D_i, the number of dishes that cow i likes and the number of drinks that cow i likes. The next F_i integers denote the dishes that cow i will eat, and the D_i integers following that denote the drinks that cow i will drink.

Output

Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

Sample Input

4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3

Sample Output

3

Hint

One way to satisfy three cows is:
Cow 1: no meal

Cow 2: Food #2, Drink #2

Cow 3: Food #1, Drink #1

Cow 4: Food #3, Drink #3

The pigeon-hole principle tells us we can do no better since there
are only three kinds of food or drink. Other test data sets are more
challenging, of course.

Source

USACO 2007 Open Gold

题意：农夫为他的 N (1 ≤ N ≤ 100) 牛准备了 F (1 ≤ F ≤ 100)种食物和 D (1 ≤ D ≤ 100) 种饮料。每头牛都有各自喜欢的食物和饮料，而每种食物或饮料只能分配给一头牛。最多能有多少头牛可以同时得到喜欢的食物和饮料？

其实这就是一个最大流 关键是建图

因为题目问的是多少头牛可以得到喜欢的食物和饮料，当时建图我就按照食物-->牛-->饮料这样建图

但存在一种牛可以满足同时得到多种自己喜欢的食物和饮料，所以这样是错误的  后来看了题解  才发现要将牛拆成两个点

食物-->牛---->牛----->饮料   流量都为1  这样就可以保证求到的一定是最大流

写了个前向星+Dinc的板子

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const double PI=acos(-1.0);
const double eps=0.0000000001;
const int INF=1e9;
const int N=+;
int vis[][];
int dis[N];
int head[N];
int a[N],b[N];
int n,f,d;
int tot;
struct node{
    int to,next,flow,c;
}edge[N<<];
void init(){
    memset(head,-,sizeof(head));
    tot=;
}
void add(int u,int v,int c){
    edge[tot].to=v;
    edge[tot].flow=c;
    edge[tot].next=head[u];
    head[u]=tot++;
    edge[tot].to=u;
    edge[tot].flow=;
    edge[tot].next=head[v];
    head[v]=tot++;

}
int BFS(int s,int t){
    queue<int>q;
    memset(dis,-,sizeof(dis));
    dis[s]=;
    q.push(s);
    while(!q.empty()){
        int x=q.front();
        q.pop();
        if(x==t)return ;
        for(int i=head[x];i!=-;i=edge[i].next){
            int v=edge[i].to;
            if(edge[i].flow&&dis[v]==-){
                dis[v]=dis[x]+;
                q.push(v);
            }
        }
    }
    if(dis[t]==-)return ;
    return ;
}
int DFS(int s,int flow){
    if(s==*n+d+f+)return flow;
    int ans=;
    for(int i=head[s];i!=-;i=edge[i].next){
        int v=edge[i].to;
        if(edge[i].flow&&dis[v]==dis[s]+){
            int f=DFS(v,min(flow-ans,edge[i].flow));
            edge[i].flow-=f;
            edge[i^].flow+=f;
            ans+=f;
            if(ans==flow)return ans;
        }
    }
    return ans;
}
int Dinc(int s,int t){
    int flow=;
    while(BFS(s,t)){
        //cout<<1<<endl;
        flow+=DFS(s,INF);
        //cout<<flow<<endl;
    }
    return flow;
}
int main(){
    while(scanf("%d%d%d",&n,&f,&d)!=EOF){
        init();
        int s=;
        for(int i=;i<=f;i++){
            add(s,*n+i,);
        }
        for(int i=;i<=d;i++){
            add(*n+f+i,*n+f+d+,);
        }
        for(int i=;i<=n;i++){
            add(i,i+n,);
        }
        int D,F;
        for(int i=;i<=n;i++){
            scanf("%d%d",&D,&F);
            for(int j=;j<=D;j++){
                scanf("%d",&a[j]);
                add(*n+a[j],i,);
            }
            for(int j=;j<=F;j++){
                scanf("%d",&b[j]);
                add(n+i,*n+f+b[j],);
            }
        }
        cout<<Dinc(s,*n+f+d+)<<endl;
    }

}