链接:

https://vjudge.net/problem/POJ-3281

题意:

农夫为他的 N (1 ≤ N ≤ 100) 牛准备了 F (1 ≤ F ≤ 100)种食物和 D (1 ≤ D ≤ 100) 种饮料。每头牛都有各自喜欢的食物和饮料，而每种食物或饮料只能分配给一头牛。最多能有多少头牛可以同时得到喜欢的食物和饮料？

思路:

建图,s->食物->牛前->牛后->饮料->汇点.
开始不知道把牛放中间,看了题解,先把食物给牛,在让牛去找饮料.给牛拆点是让牛只能选一个.
在跑最大流即可.

代码:

#include <iostream>
#include <cstdio>
#include <vector>
#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
using namespace std;
typedef long long LL;

const int MAXN = 1e2+10;
const int INF = 1e9;

struct Edge
{
    int from, to, cap;
    Edge(int f, int t, int c):from(f), to(t), cap(c){};
};

vector<int> G[MAXN*5];
vector<Edge> edges;
int Dgree[MAXN*5];
int f, d, n, s, t;

void AddEdge(int from, int to, int cap)
{
    edges.push_back(Edge(from, to, cap));
    edges.push_back(Edge(to, from, 0));
    G[from].push_back(edges.size()-2);
    G[to].push_back(edges.size()-1);
}

bool Bfs()
{
    memset(Dgree, -1, sizeof(Dgree));
    queue<int> que;
    que.push(s);
    Dgree[s] = 0;
    while (!que.empty())
    {
        int u = que.front();
        que.pop();
        for (int i = 0;i < G[u].size();i++)
        {
            Edge &e = edges[G[u][i]];
            if (e.cap > 0 && Dgree[e.to] == -1)
            {
                Dgree[e.to] = Dgree[u]+1;
                que.push(e.to);
            }
        }
    }
    return Dgree[t] != -1;
}

int Dfs(int u, int flow)
{
    if (u == t)
        return flow;
    int res = 0;
    for (int i = 0;i < G[u].size();i++)
    {
        Edge &e = edges[G[u][i]];
        if (e.cap > 0 && Dgree[e.to] == Dgree[u]+1)
        {
            int tmp = Dfs(e.to, min(flow, e.cap));
            res += tmp;
            e.cap -= tmp;
            flow -= tmp;
            edges[G[u][i]^1].cap += tmp;
            if (flow == 0)
                break;
        }
    }
    if (res == 0)
        Dgree[u] = -1;
    return res;
}

int Dinic()
{
    int res = 0;
    while (Bfs())
        res += Dfs(s, INF);
    return res;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //n头牛,每头两个编号i*2-1, i*2
    //食物编号,2*n+i
    //水编号,2*n+f+i
    //源点s = 0, t = 2*n+f+d+1
    while (cin >> n >> f >> d)
    {
        s = 0, t = 2*n+f+d+1;
        for (int i = s;i <= t;i++)
            G[i].clear();
        edges.clear();
        for (int i = 1;i <= n;i++)
        {
            AddEdge(i*2-1, i*2, 1);
            int ff, dd, node;
            cin >> ff >> dd;
            while (ff--)
            {
                cin >> node;
                AddEdge(2*n+node, i*2-1, 1);
            }
            while (dd--)
            {
                cin >> node;
                AddEdge(2*i, 2*n+f+node, 1);
            }
        }
        for (int i = 1;i <= f;i++)
            AddEdge(0, 2*n+i, 1);
        for (int i = 1;i <= d;i++)
            AddEdge(2*n+f+i, t, 1);
        int res = Dinic();
        cout << res << endl;
    }

    return 0;
}

标签：int,res,cap,Dining,edges,Dgree,POJ,include,3281
来源： https://www.cnblogs.com/YDDDD/p/11298003.html