POJ 3281 Dining 网络流最大流

B - Dining
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88038#problem/B

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_iintegers 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

题意

有n头牛，每头牛有f种喜欢的食物和d种喜欢的饮料，然后问你，究竟多有多少头牛，可以吃到满意的食物和饮料

题解：

s-食物-牛-牛-饮料

因为牛只能用一次，所以注意拆点

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 5000
#define mod 10007
#define eps 1e-9
int Num;
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************
namespace NetFlow
{
    const int MAXN=,MAXM=,inf=1e9;
    struct Edge
    {
        int v,c,f,nx;
        Edge() {}
        Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {}
    } E[MAXM];
    int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz;
    void init(int _n)
    {
        N=_n,sz=; memset(G,-,sizeof(G[])*N);
    }
    void link(int u,int v,int c)
    {
        E[sz]=Edge(v,c,,G[u]); G[u]=sz++;
        E[sz]=Edge(u,,,G[v]); G[v]=sz++;
    }
    int ISAP(int S,int T)
    {//S -> T
        int maxflow=,aug=inf,flag=false,u,v;
        for (int i=;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=;
        for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false)
        {
            for (int &it=cur[u];~it;it=E[it].nx)
            {
                if (E[it].c>E[it].f&&dis[u]==dis[v=E[it].v]+)
                {
                    if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f;
                    pre[v]=u,u=v; flag=true;
                    if (u==T)
                    {
                        for (maxflow+=aug;u!=S;)
                        {
                            E[cur[u=pre[u]]].f+=aug;
                            E[cur[u]^].f-=aug;
                        }
                        aug=inf;
                    }
                    break;
                }
            }
            if (flag) continue;
            int mx=N;
            for (int it=G[u];~it;it=E[it].nx)
            {
                if (E[it].c>E[it].f&&dis[E[it].v]<mx)
                {
                    mx=dis[E[it].v]; cur[u]=it;
                }
            }
            if ((--gap[dis[u]])==) break;
            ++gap[dis[u]=mx+]; u=pre[u];
        }
        return maxflow;
    }
    bool bfs(int S,int T)
    {
        static int Q[MAXN]; memset(dis,-,sizeof(dis[])*N);
        dis[S]=; Q[]=S;
        for (int h=,t=,u,v,it;h<t;++h)
        {
            for (u=Q[h],it=G[u];~it;it=E[it].nx)
            {
                if (dis[v=E[it].v]==-&&E[it].c>E[it].f)
                {
                    dis[v]=dis[u]+; Q[t++]=v;
                }
            }
        }
        return dis[T]!=-;
    }
    int dfs(int u,int T,int low)
    {
        if (u==T) return low;
        int ret=,tmp,v;
        for (int &it=cur[u];~it&&ret<low;it=E[it].nx)
        {
            if (dis[v=E[it].v]==dis[u]+&&E[it].c>E[it].f)
            {
                if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f)))
                {
                    ret+=tmp; E[it].f+=tmp; E[it^].f-=tmp;
                }
            }
        }
        if (!ret) dis[u]=-; return ret;
    }
    int dinic(int S,int T)
    {
        int maxflow=,tmp;
        while (bfs(S,T))
        {
            memcpy(cur,G,sizeof(G[])*N);
            while (tmp=dfs(S,T,inf)) maxflow+=tmp;
        }
        return maxflow;
    }
}
using namespace NetFlow;
map<pair<int,int> ,int> H;
int tot=;
int get_id(int x,int y)
{
    pair<int,int> A;
    A.first = x;
    A.second = y;
    if(H[A]!=)
        return H[A];
    else H[A]=tot++;
    return H[A];
}

int main()
{
    init();
    int n=read(),f=read(),d=read();
    for(int i=;i<=f;i++)
        link(get_id(,),get_id(i,),);
    for(int i=;i<=d;i++)
        link(get_id(i,),get_id(,),);
    for(int i=;i<=n;i++)
        link(get_id(i,),get_id(i,),);
    for(int i=;i<=n;i++)
    {
        int ff=read(),dd=read();
        for(int j=;j<=ff;j++)
        {
            int x=read();
            link(get_id(x,),get_id(i,),);
        }
        for(int j=;j<=dd;j++)
        {
            int x=read();
            link(get_id(i,),get_id(x,),);
        }
    }
    printf("%d\n",dinic(get_id(,),get_id(,)));
}