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

Factors and Multiples

Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu

SubmitStatus

Description

You will be given two sets of integers. Let's call them set A and setB. Set A contains n elements and setB contains m elements. You have to remove k₁ elements from set A and k₂ elements from setB so that of the remaining values no integer in set B is a multiple of any integer in setA. 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 ofQ 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 integers6, 7 or 9 is a multiple of 4 or5.

So for this case the answer is 3 (two from setA 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 byn positive integers. The second line starts with m followed bym 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;}