lightoj1149 - Factors and Multiples【二分图最大匹配】

1149 - Factors and Multiples

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Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for Sample Input

2

4 2 3 4 5

4 6 7 8 9

3 100 200 300

1 150

Case 1: 3

Case 2: 0

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#define inf 0x3f3f3f3fusing namespace std;const int maxn=110;int n,m;int p[maxn];int numa[maxn];int numb[maxn];bool vis[maxn];bool find(int k){for(int i=0;i<m;++i){if(!vis[i]&&numb[i]%numa[k]==0){vis[i]=true;if(p[i]==-1||find(p[i])){p[i]=k;return true;}}}return false;}int main(){int i,j,k=1,t;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;++i){scanf("%d",&numa[i]);}scanf("%d",&m);for(i=0;i<m;++i){scanf("%d",&numb[i]);}int ans=0;memset(p,-1,sizeof(p));for(i=0;i<n;++i){memset(vis,false,sizeof(vis));if(find(i))ans++;}printf("Case %d: %d\n",k++,ans);}return 0;}