POJ 3281 Dining

Dining

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3281
64-bit integer IO format: %lld      Java class name: Main

 

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 Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers  denote the dishes that cow i will eat, and the Di 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

Source

USACO 2007 Open Gold

 

解题：最大流，题目貌似要指定牛要么不配，要配就要配全，Food+Drink都要有，把牛拆了。。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 500;
struct arc{
    int to,flow,next;
    arc(int x = 0,int y = 0,int z = -1){
        to = x;
        flow = y;
        next = z;
    }
};
arc e[200000];
int head[maxn],d[maxn],cur[maxn];
int tot,N,F,D,s,t;
void add(int u,int v,int flow){
    e[tot] = arc(v,flow,head[u]);
    head[u] = tot++;
    e[tot] = arc(u,0,head[v]);
    head[v] = tot++;
}
bool bfs(){
    memset(d,-1,sizeof(d));
    queue<int>q;
    d[s] = 1;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = e[i].next){
            if(e[i].flow && d[e[i].to] == -1){
                d[e[i].to] = d[u] + 1;
                q.push(e[i].to);
            }
        }
    }
    return d[t] > -1;
}
int dfs(int u,int low){
    if(u == t) return low;
    int tmp = 0,a;
    for(int &i = cur[u]; ~i; i = e[i].next){
        if(e[i].flow && d[e[i].to] == d[u] + 1 &&(a=dfs(e[i].to,min(low,e[i].flow)))){
            e[i].flow -= a;
            e[i^1].flow += a;
            tmp += a;
            low -= a;
            if(!low) break;
        }
    }
    if(!tmp) d[u] = -1;
    return tmp;
}
int dinic(){
    int ans = 0;
    while(bfs()){
        memcpy(cur,head,sizeof(head));
        ans += dfs(s,INF);
    }
    return ans;
}
int main() {
    while(~scanf("%d %d %d",&N,&F,&D)){
        memset(head,-1,sizeof(head));
        s = tot = 0;
        int n = F + D,a,b,c;
        t = n + (N<<1|1);
        for(int i = 1; i <= N; ++i){
            scanf("%d %d",&a,&b);
            while(a--){
                scanf("%d",&c);
                add(c,n+(i<<1)-1,1);
            }
            add(n+(i<<1)-1,n+(i<<1),1);
            while(b--){
                scanf("%d",&c);
                add(n+(i<<1),F+c,1);
            }
        }
        for(int i = 1; i <= F; ++i) add(s,i,1);
        for(int i = 1; i <= D; ++i) add(F+i,t,1);
        printf("%d
",dinic());
    }
    return 0;
}