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自己选择的路,就算跪着也要走完

01/24
20:37
图论 网络流

POJ 3281 Dining 网络流-最大流

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

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

Soluiton

网络流的裸题,敲个Dinic存个板子。 顺便说一下Dinic。 Dinic多路增广,还用到了阻塞流?嗯对……
Dinic的优化:当前弧优化,以及分层优化(针对有环的情况)

关于这道题,给定n头牛,f种食物,d种饮料,每一头牛有喜欢其中的几种饮料和食物,但是每种食物和饮料只有一个。问能最大满足多少头牛同时吃到喜欢的食物,喝到喜欢的饮料。
很容易想到建两个源汇点s,t,s向每种食物/饮料连边,每种食物/饮料向对应的牛连边,牛再向对应的饮料/食物连边,然后每种饮料/食物向t连边,流量都是1. 可是这样的话,一头牛有可能吃到多种喜欢的食物,喝到多种喜欢的饮料,显然不能得到最优解。怎么处理呢?把每头牛拆成两个点,两个点之间连一条流量为1的边,这样就处理好了。