Loj 1149 二分图最大匹配

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. k1should 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

 

---

PROBLEM SETTER: SOHEL HAFIZ

SPECIAL THANKS: JANE ALAM JAN

 // 

题目大意：

集合A中有N个数，集合B中有M个数，现在需要在A集合中去掉K1个数，在B集合中去掉K2个数，现在需要（K1+K2）尽可能的小，保证剩余部分中,B集合中的任意数都不能整除A集合中的任意数。

对于B[i]%A[i]==0的两个建连接，然后求最大二分匹配==最小点覆盖==最大二分匹配数。

This is the code :

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define LL long long
#define ULL unsigned long long     //1844674407370955161
#define INT_INF 0x3f3f3f3f      //1061109567
#define LL_INF 0x3f3f3f3f3f3f3f3f //4557430888798830399
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。

inline int read()//输入外挂
{
    int ret=0, flag=0;
    char ch;
    if((ch=getchar())=='-')
        flag=1;
    else if(ch>='0'&&ch<='9')
        ret = ch - '0';
    while((ch=getchar())>='0'&&ch<='9')
        ret=ret*10+(ch-'0');
    return flag ? -ret : ret;
}

struct Max_Match
{
    int p,q;            //左右两点集的点个数
    vector<int> g[110]; //邻接矩阵,g[i]中保存的是左边第i个节点所连的右边节点的编号
    int left[310];      //left[i]==j表右边第i个点与左边第j个点匹配,为-1表无点匹配
    bool vis[310];      //表右边第i个点是否已经被访问过

    void init(int p,int q)
    {
        this->p=p;
        this->q=q;
        for(int i=0; i<=p; i++)
            g[i].clear();
        memset(left,-1,sizeof(left));
    }

    bool match(int u)
    {
        for(int j=0; j<g[u].size(); j++)
        {
            int v=g[u][j];
            if(!vis[v])
            {
                vis[v]=true;
                if(left[v]==-1 || match(left[v]) )
                {
                    left[v]=u;//此步相当于将增广路取反
                    return true;
                }
            }
        }
        return false;
    }

    void solve()
    {
        int ans=0;//记录匹配边数
        for(int i=1; i<=p; i++)
        {
            memset(vis,0,sizeof(vis));
            if(match(i))
                ans++;
        }
        printf("%d\n",ans);
    }
} MM;

int a[105];
int b[105];

int main()
{
    int T=read();
    for(int t=1;t<=T;++t)
    {
        int p=read();
        for(int i=1;i<=p;++i)
            a[i]=read();
        int q=read();
        MM.init(q,p); //注意哪个变量是左边集合
        for(int i=1;i<=q;++i)
        {
            b[i]=read();
            for(int j=1;j<=p;++j)
            {
                if(b[i]%a[j]==0)
                    MM.g[i].push_back(j);
            }
        }
        printf("Case %d: ",t);
        MM.solve();
    }
    return 0;
}