poj3281—Dining（最大流）

题目链接：传送门

Dining

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 18284 Accepted: 8155

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 32 2 1 2 3 12 2 2 3 1 22 2 1 3 1 22 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.

解题思路：建图s—食物—牛—牛—饮料—t，权值为1，相同的牛之间建一条边确保了每头牛只选择一种搭配，然后在图上跑个最大流。

#include <cstdio>#include <algorithm>#include <queue>#include <iostream>using namespace std;const int N = 420;const int M = 20900;//按s——食物——牛——牛——饮料——t建图，边权都为1//每头牛都只能选一种搭配，所以同一头牛间建条边//边的结构struct edge_t{    int node;    int c;//c为容量    edge_t* next;    edge_t* redge;//指向反向边}Edge[M*2];int ECnt;//图的邻接表edge_t* Ver[N];void init(){    ECnt = 0;    fill(Ver,Ver+N,(edge_t*)0);}//生成双向边void mkEdge(int a,int b,int c){    int t1 = ECnt++;    int t2 = ECnt++;    Edge[t1].node = b;    Edge[t1].c = c;    Edge[t1].next = Ver[a];    Edge[t1].redge = Edge + t2;    Ver[a] = Edge + t1;    Edge[t2].node = a;    Edge[t2].c = 0;    Edge[t2].next = Ver[b];    Edge[t2].redge = Edge + t1;    Ver[b] = Edge + t2;}int L[N];//层次图//建立残留网络从源s到汇t的层次图bool bfs(int s,int t){    fill(L,L+N,-1);    queue<int>q;    q.push(s);    L[s] = 0;    while( !q.empty() ){        int u = q.front();        q.pop();        //寻找还有残量的边        for(edge_t*p=Ver[u];p;p=p->next){            if ( p->c <= 0 ) continue;            int v = p->node;            if ( -1 != L[v] ) continue;            q.push(v);            L[v] = L[u] + 1;        }    }    return -1 != L[t];}//在层次图上搜索增广路径，本质上就是搜索可以增广的流量//这个流量是各层之间流量的最小值//u为当前节点，cf为当前层的最小流，t为汇点int dfs(int u,int e,int cf){    if ( u == e ) return cf;    int tf = 0;  //tf记录u往下一层的总可行流量    for(edge_t*p=Ver[u];p;p=p->next){        int v = p->node;        int c = p->c;        if ( L[u] + 1 == L[v] && c > 0 && cf > tf ){            int f = dfs(v,e,min(c,cf-tf));            if ( 0 == f ) continue;            p->c -= f;//正向边减去可行流量            p->redge->c += f;//反向边加上            tf += f;        }    }    if ( 0 == tf ) L[u] = -1;//修改层次图    return tf;}//Dinic算法，s为源，t为汇int Dinic(int s,int t){    int ret = 0;    while( bfs(s,t) ){//第一步建立分层图        int ans;        //第二步在分层图上查找一条增广路径的可行流量        while( ans = dfs(s,t,INT_MAX) )            ret += ans;    }    return ret;}//n头牛，f中食物，d种喝的void Build( int n , int f , int d ){    int a,b,c;    //源点和食物建边    for( int i = 1 ; i <= f ; ++i ) mkEdge(0,i+2*n,1);    //饮料喝汇点建边    for( int i = 1 ; i <= d ; ++i ) mkEdge(2*n+f+i,2*n+f+d+1,1);    for( int i = 1 ; i <= n ; ++i ){        scanf("%d%d",&a,&b);        //食物和牛建边        for( int j = 0 ; j < a ; ++j){            scanf("%d",&c);            mkEdge(c+2*n,i,1);        }        //牛和牛建边        mkEdge(i,n+i,1);        //牛和饮料建边        for( int j = 0 ; j < b ; ++j ){            scanf("%d",&c);            mkEdge(n+i,2*n+f+c,1);        }    }}int main(){    int n,f,d;    while( ~scanf("%d%d%d",&n,&f,&d) ){        init();        Build(n,f,d);        printf("%d\n",Dinic(0,2*n+f+d+1));    }    return 0;}