light oj 1149 - Factors and Multiples （二分匹配）

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

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

 

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PROBLEM SETTER: SOHEL HAFIZ

SPECIAL THANKS: JANE ALAM JAN

题目大意： 主要是看图片上的那个小孩，看看图片，看看样例，大概就是二分匹配的题； 输入两组数，

                     第二组的数是第一组数任意一个的倍数，表明两者有关系，比如样例中的 6 是2、3 的倍数，

                     表明有关系，求最大的匹配数 

解析：         基本二分匹配

代码如下：

#include<iostream>#include<algorithm>#include<map>#include<string>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#define N  1009using namespace std;const int inf = 0x3f3f3f3f;const int mod = 1000003;int mp[N][N], used[N], match[N], m, n;int Find(int u){    for(int i = 1; i <= n; i++)    {        if(!used[i] && mp[u][i])        {            used[i] = 1;            if(match[i] == -1 || Find(match[i]))            {                match[i] = u;                return 1;            }        }    }    return 0;}int main(){    int a[N], b[N];    int t, cnt = 0, i, j;    cin >> t;    while(t--)    {        scanf("%d", &n);        for(i = 1; i <= n; i++)        {            scanf("%d", &a[i]);        }        scanf("%d", &m);        for(i = 1; i <= m; i++)        {            scanf("%d", &b[i]);        }        memset(mp, 0, sizeof(mp));        memset(match, -1, sizeof(match));        for(i = 1; i <= m; i++)            for(j = 1; j <= n; j++)            if(b[i] % a[j] == 0)            mp[i][j] = 1;        int ans = 0;        for(i = 1; i <= m; i++)        {            memset(used, 0, sizeof(used));            ans += Find(i);        }        printf("Case %d: %d\n", ++cnt, ans);    }}