B - Dining POJ - 3281 网络流

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

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.

 

 

 

 

牛吃草问题，这个牛是一个点，有因为每一头牛只能用一次，所以要进行拆点。

这个图很好建的，我就不说了。

 

#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cstring>
#include <vector>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1e5 + ;
int n, f, d;
struct node
{
    int from, to, cap, flow;
    node(int from = , int to = , int cap = , int flow = ) :from(from), to(to), cap(cap), flow(flow) {}
};
vector<node>e;
vector<int>G[maxn];
int level[maxn], iter[maxn];
void add(int u, int v, int w)
{
    e.push_back(node(u, v, w, ));
    e.push_back(node(v, u, , ));
    int m = e.size();
    G[u].push_back(m - );
    G[v].push_back(m - );
}

void bfs(int s)//这个是为了构建层次网络，也就是level的构建
{
    memset(level, -, sizeof(level));
    queue<int>que;
    que.push(s);
    level[s] = ;
    while (!que.empty())
    {
        int u = que.front(); que.pop();
        for (int i = ; i < G[u].size(); i++)
        {
            node &now = e[G[u][i]];
            if (now.cap > now.flow&&level[now.to] < )//只有这个没有满并且没有被访问过才可以被访问
            {
                level[now.to] = level[u] + ;
                que.push(now.to);
            }
        }
    }
}

int dfs(int u, int v, int f)
{
    if (u == v) return f;
    for (int &i = iter[u]; i < G[u].size(); i++)
    {
        node &now = e[G[u][i]];
        if (now.cap > now.flow&&level[now.to] > level[u])
        {
            int d = dfs(now.to, v, min(f, now.cap - now.flow));
            if (d > )
            {
                now.flow += d;
                e[G[u][i] ^ ].flow -= d;
                return d;
            }
        }
    }
    return ;
}

int Maxflow(int s, int t)
{
    int flow = ;
    while ()
    {
        bfs(s);
        if (level[t] < ) return flow;
        memset(iter, , sizeof(iter));
        int f;
        while ((f = dfs(s, t, inf) > )) flow += f;
    }
}
void init()
{
    for (int i = ; i <= n + ; i++) G[i].clear();
    e.clear();
}

int main()
{
    while (cin >> n >> f >> d)
    {
        init();
        int s = , t = f +  * n + d + ;
        for (int i = ; i <= f; i++) add(s, i, );
        for (int i = ; i <= n; i++)
        {
            int a, b;
            cin >> a >> b;
            while (a--)//与牛i相连
            {
                int x;
                cin >> x;
                add(x, f + i, );
            }
            add(f + i, f + n + i, );
            while (b--)
            {
                int x;
                cin >> x;
                add(f + n + i, f +  * n + x, );
            }
        }
        for (int i = ; i <= d; i++) add(f +  * n + i, t, );
        int ans = Maxflow(s, t);
        cout << ans << endl;
    }
    return ;
}