POJ3281--Dining

本文介绍了一种解决农场中牛群特定餐饮偏好的最优分配问题的算法。通过使用增广路径算法,该方案能够最大化满足牛群对食物和饮料偏好组合的数量。文章详细展示了输入输出格式及示例,并提供了一个具体的实现代码。

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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: NF, 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
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 508
#define edge 50080
#define inf 0x3f3f3f3f
int first[maxn],dis[maxn],num[maxn];
int vv[edge],ww[edge],nxt[edge];
int e,NN;
inline int min(int a,int b)
{
	return a>b?b:a;
}
void addEdge(int u,int v,int w)
{
	vv[e] = v;	ww[e] = w;		nxt[e] = first[u];	first[u] = e++;
	vv[e] = u;	ww[e] = 0;		nxt[e] = first[v];		first[v] = e++;
}
int dfs(int u,int s,int d,int cost)
{
	if(u == d)	return cost;
	int ans = 0;
	int _min = NN;
	for(int i = first[u];i != -1;i = nxt[i])
	{
		int v = vv[i];
		if(ww[i])
		{
			if(dis[v] + 1 == dis[u])
			{
				int t = dfs(v,s,d,min(ww[i],cost));
				ww[i] -= t;
				ww[i^1] += t;
				ans += t;
				cost -= t;
				if(dis[s] == NN)		return ans;
				if(!cost)	break;
			}
			if(_min > dis[v])		_min = dis[v];
		}
	}
	if(!ans)
	{
		if(--num[dis[u]] == 0)	dis[s] = NN;
		dis[u] = _min + 1;
		++num[dis[u]];
	}
	return ans;
}
int isap(int s,int d)
{
	memset(dis,0,sizeof(dis));
	memset(num,0,sizeof(num));
	num[0] = NN;
	int ans = 0;
	while(dis[s] < NN)
		ans += dfs(s,s,d,inf);
	return ans;
}
int main()
{
	//freopen("in.txt","r",stdin);
	int n,f,d;
	while(scanf("%d%d%d",&n,&f,&d)==3)
	{
		memset(first,-1,sizeof(first));
		e = 0;
		NN = 2+f+d+2*n;
		for(int i = 1;i <= f;i++)
		{	
			addEdge(0,i,1);
		}
		for(int i = f+2*n+1;i <= f+2*n+d;i++)
		{
			addEdge(i,f+2*n+d+1,1);
		}
		for(int i = f+1;i <= f+n;i++)
		{
			addEdge(i,i+n,1);
		}
		for(int i = 1;i <= n;i++)
		{
			int ff,dd;		scanf("%d%d",&ff,&dd);
			for(int j = 1;j <= ff;j++)
			{
				int u;	scanf("%d",&u);
				addEdge(u,f+i,1);
			}
			for(int j = 1;j <= dd;j++)
			{
				int v;	scanf("%d",&v);
				addEdge(f+i+n,f+2*n+v,1);
			}
		}
		printf("%d\n",isap(0,f+2*n+d+1));
	}
	return 0;
}


 

 

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