poj 3281 Dining Maxflow

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http://acm.pku.edu.cn/JudgeOnline/problem?id=3281

poj 3281 Dining
Time Limit:2000MS Memory Limit:65536K
Total Submit:308 Accepted:120

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

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

Description:
N头牛，f种食物，d种饮料，每只牛ci有喜欢吃的fi种食物和喜欢喝的di种饮料，要求最多有多少牛能吃到自己喜欢的食物和喝到自己喜欢的饮料
这是一道网络流很好的练习题，还贯穿着最大匹配的思想，只是建图稍复杂。

Solution:
具体建图的方法是，把食物和饮料放在牛的两边，再把牛分成两个结点，两个分开的牛结点之间用一条容量为1的边相连，分别建源点汇点，源点到食物的每条边容量为1，汇点到饮料的每条边容量为1，牛和食物饮料之间的边容量也为1需要注意，对于每头牛的结点有容量的限制，限制容量为1，因为一头牛只能吃一种食物，喝一种饮料
用Edmonds－Karp求最大流