<每日一题>Day 9：POJ-3281.Dining(拆点 + 多源多汇+ 网络流 )

Dining

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24945		Accepted: 10985

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

　　都在代码里了，干/drink.jpg

/*
    POJ 3281 最大流 + 拆点
    源点 -> food -> 牛左 -> 牛右 -> Drink -> 汇点
    建图时注意将上面的所有边的容量设置为1，这样就可以保证一头牛
    只吃一种食物喝一种饮料，转化之后肯定就知道是最大流了

    拆点技巧：为了保证同一个东西满足两个条件，则将其拆分为两个
    公共边的点分别进行求解。

    嘤嘤嘤，为什么做完之后感觉这个题其实建图也是很好想的，就是拆个点，原谅自己太差
    唉，都是幻觉
*/

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn=+, INF = 0x3f3f3f3f;
struct Edge
{
    Edge(){}
    Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){}
    int from,to,cap,flow;
};

struct Dinic
{
    int n,m,s,t;            //结点数,边数(包括反向弧),源点与汇点编号
    vector<Edge> edges;     //边表 edges[e]和edges[e^1]互为反向弧
    vector<int> G[maxn];    //邻接表,G[i][j]表示结点i的第j条边在e数组中的序号
    bool vis[maxn];         //BFS使用,标记一个节点是否被遍历过
    int d[maxn];            //从起点到i点的距离
    int cur[maxn];          //当前弧下标

    void init(int n,int s,int t)
    {
        this->n=n,this->s=s,this->t=t;
        for(int i=;i<=n;i++) G[i].clear();
        edges.clear();
    }

    void AddEdge(int from,int to,int cap)
    {
        edges.push_back( Edge(from,to,cap,) );
        edges.push_back( Edge(to,from,,) );
        m = edges.size();
        G[from].push_back(m-);
        G[to].push_back(m-);
    }

    bool BFS()
    {
        memset(vis,false,sizeof(vis));
        queue<int> Q;//用来保存节点编号的
        Q.push(s);
        d[s]=;
        vis[s]=true;
        while(!Q.empty())
        {
            int x=Q.front(); Q.pop();
            for(int i=; i<G[x].size(); i++)
            {
                Edge& e=edges[G[x][i]];
                if(!vis[e.to] && e.cap>e.flow)
                {
                    vis[e.to]=true;
                    d[e.to] = d[x]+;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x,int a)
    {
        if(x==t || a==)return a;
        int flow=,f;//flow用来记录从x到t的最小残量
        for(int& i=cur[x]; i<G[x].size(); i++)
        {
            Edge& e=edges[G[x][i]];
            if(d[x]+==d[e.to] && (f=DFS( e.to,min(a,e.cap-e.flow) ) )> )
            {
                e.flow +=f;
                edges[G[x][i]^].flow -=f;
                flow += f;
                a -= f;
                if(a==) break;
            }
        }
        return flow;
    }

    int Maxflow()
    {
        int flow=;
        while(BFS())
        {
            memset(cur,,sizeof(cur));
            flow += DFS(s,INF);
        }
        return flow;
    }
}Dinic;

int main() {
    int n, f, a, b, c, d;
    scanf("%d %d %d", &n, &f, &d);
    int s = n *  + f + d + , t = s + ;
    Dinic.init(n, s, t);
    for(int i = ; i <= f; i ++) {//n * 2 - 1 ~ n * 2 + f - 1存储从s -> 食物的边
        Dinic.AddEdge(s, n *  + i, );
    }
    for(int i = ; i <= d; i ++) {//n * 2 + f ~ n * 2 + f + d - 1存储从饮料 -> t的边
        Dinic.AddEdge(n *  + f + i, t, );
    }
    for(int i = ; i <= n; i ++) {
        Dinic.AddEdge(i ,n + i, );//牛拆点之后的建立的边1 ~ 2 * n
        scanf("%d %d", &a, &b);
        for(int j = ; j < a; j ++) {
            scanf("%d", &c);
            Dinic.AddEdge(n *  + c, i, );//食物与牛建边
        }
        for(int j = ; j < b; j ++) {
            scanf("%d", &c);
            Dinic.AddEdge(i + n, n *  + f + c, );//饮料与牛建边
        }
    }
    printf("%d\n", Dinic.Maxflow());//模版最大流？
    return ;
}