（LightOJ 1149） Factors and Multiples

题目链接：http://lightoj.com/volume_showproblem.php?problem=1149

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 {, , , } and set B is {, , , }. By removing  and  from A and  from B, we get the sets {, } and {, , }. Here none of the integers ,  or  is a multiple of  or .

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

Input
Input starts with an integer T (≤ ), 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 [, ]. Each element of the two sets will fit in a  bit signed integer.

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

Sample Input

Sample Output
Case :
Case : 

方法：二分匹配，求最大匹配数

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <math.h>
#include <algorithm>
#include <queue>
using namespace std;

#define met(a,b) memset(a,b,sizeof(a))
#define ll long long
#define N 505
int Map[N][N],vis[N],used[N];
int a[N],b[N];
int n,m;
int han(int u)
{
    for(int i=;i<=m;i++)
    {
        if(!vis[i] && Map[u][i])
        {
            vis[i]=;
            if(!used[i] || han(used[i]))
            {
                used[i]=u;
                return ;
            }
        }
    }
    return ;
}
int main()
{
    int t,con=;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);met(Map,);
        for(int i=;i<=n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        for(int i=;i<=m;i++)
        scanf("%d",&b[i]);
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=m;j++)
                if(b[j]%a[i]==)
                Map[i][j]=;
        }
        met(used,);int sum=;
        for(int i=;i<=n;i++)
        {
            met(vis,);
            sum+=han(i);
        }

        printf("Case %d: %d\n",con++,sum);
    }
    return ;
}