POJ 3281 -- Dining（最大流，拆点建图）

题目链接

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 Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers denote the dishes that cow i will eat, and the Di 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

AC

建图思路：
题目问的是最多有几头牛可以同时得到自己想要的食物和水，如果一头牛可以同时可以得到两对合适的食物和水，这样就浪费了，所以要把牛拆开，牛和自己建一条权值为1的边，这样保证一头牛如果可以的话只能得到一对食物和水源，从而使得越多的牛可以得到对应的食物和水
1. 源点和所有食物建立一条权值为1的边
2. 食物和牛建立权值为1的边
3. 牛的副本和对应的水建立权值为1的边
4. 水和汇点建立权值为1的边
5. 牛和牛的副本建立权值为1的边
建好图之后，跑一遍最大流，得到的就是最多牛的个数

#include <iostream>
#include <stdio.h>
#include <map>
#include <vector>
#include <queue>
#include <algorithm>
#include <cmath>
#define N 300005
#include <cstring>
#define ll long long
#define P pair<int, int>
#define mk make_pair
using namespace std;int a[405][405], pre[405];
bool vis[405];
int inf = 0x3f3f3f3f;
int n, f, d;
// EK模板
bool bfs(int s, int e) {memset(pre, -1, sizeof(pre));memset(vis, false, sizeof(vis));vis[s] = true;pre[s] = s;queue<int> que;que.push(s);while (!que.empty()) {int t = que.front();que.pop();for (int i = 0; i <= n + n + f + d + 1; ++i) {if (vis[i] || a[t][i] == 0) continue;vis[i] = true;pre[i] = t;if (i == e) return true;que.push(i);}}return false;
}ll solve(int s, int e) {ll sum = 0;while (bfs(s, e)) {int MIN = inf;for (int i = e; i != s; i = pre[i]) {MIN = min(MIN, a[pre[i]][i]);}for (int i = e; i != s; i = pre[i]) {a[pre[i]][i] -= MIN;a[i][pre[i]] += MIN;}   sum += MIN;}return sum;
}int main() {
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);
#endifscanf("%d%d%d" ,&n, &f, &d);for (int i = 1; i <= n; ++i) {int l, r, t;scanf("%d%d",&l, &r);// food 和 cow 建边 for (int j = 0; j < l; ++j) {scanf("%d", &t);a[t][f + i] = 1;  }// cow 和 water 建边 for (int j = 0; j < r; ++j) {scanf("%d", &t);a[f + i + n][t + f + n + n] = 1;}}// 源点 和 食物建边 for (int i = 1; i <= f; ++i) {a[0][i] = 1;}// 水 和 汇点建边 for (int i = 1; i <= d; ++i) {a[i + f + n + n][n + f + d + 1 + n] = 1;}// 牛和牛之间建边 for (int i = 1; i <= n; ++i) {a[f + i][f + n + i] = 1;}// 跑一边EK ll ans = solve(0, n + n + f + d + 1);printf("%d\n", ans);return 0;
}