LightOJ--1149--Factors and Multiples（二分图好题）

Factors and Multiples

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

Description

You will be given two sets of integers. Let's call them set A and set B. Set A contains n elements and set B contains m elements. You have to remove k₁ elements from set A and k₂ elements from set B so that of the remaining values no integer in set B is a multiple of any integer in set A. k₁ should be in the range [0, n] and k₂ in the range [0, m].

You have to find the value of (k₁ + k₂) such that (k₁ + k₂) is as low as possible. P is a multiple of Q if there is some integer K such that P = K * Q.

Suppose set A is {2, 3, 4, 5} and set B is {6, 7, 8, 9}. By removing 2 and 3 from A and 8 from B, we get the sets {4, 5} and {6, 7, 9}. Here none of the integers 6, 7 or 9 is a multiple of 4 or 5.

So for this case the answer is 3 (two from set A and one from set B).

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

The first line of each case starts with an integer n followed by n positive integers. The second line starts with m followed by m positive integers. Both n and m will be in the range [1, 100]. Each element of the two sets will fit in a 32 bit signed integer.

Output

For each case of input, print the case number and the result.

Sample Input

2

4 2 3 4 5

4 6 7 8 9

3 100 200 300

1 150

Sample Output

Case 1: 3

Case 2: 0

Source

Problem Setter: Sohel Hafiz

Special Thanks: Jane Alam Jan

给了两个集合A，B，分别有n，m个数，从A取k1个数，B取k2个数，使得b[ j ]%a[ i ]==0的情况不存在
刚开始以为可以暴力的，但是后来发现暴力真的是挺麻烦，把图画出来之后会发现，其实就是最小点覆盖，二分图性质：最小点覆盖=最大匹配，匈牙利算法跑一次

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>map[200];
int used[200],pipei[200],a[200],b[200];
int n,m;
int find(int x)
{
	
	for(int i=0;i<map[x].size();i++)
	{
		int y=map[x][i];
		if(!used[y])
		{
			used[y]=1;
			if(pipei[y]==-1||find(pipei[y]))
			{
				pipei[y]=x;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int t,k=1;
	scanf("%d",&t);
	while(t--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(pipei,-1,sizeof(pipei));
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			map[i].clear();
		}
		scanf("%d",&m);
		for(int i=0;i<m;i++)
		scanf("%d",&b[i]);
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(b[j]%a[i]==0)
				{
					map[i].push_back(j);
				}
			}
		}
		int sum=0;
		for(int i=0;i<n;i++)
		{
			memset(used,0,sizeof(used));
			sum+=find(i);
		}
		printf("Case %d: %d\n",k++,sum);
	}
	return 0;
}