light oj 1149 Factors and Multiples(二分匹配)

LightOJ1149 :Factors and Multiples

时间限制:2000MS    内存限制:32768KByte   64位IO格式:%lld & %llu

描述

You will be given two sets of integers. Let's call them set Aand set B. Set A contains n elements and set Bcontains 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 k₂in the range [0, m].

You have to find the value of (k₁ + k₂)such that (k₁ + k₂) is as low as possible. Pis 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 Bis {6, 7, 8, 9}. By removing 2 and 3 from A and 8from 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 setA and one from set B).

输入

Input starts with an integer T (≤ 50), denoting the number of test cases.

The first line of each case starts with an integer nfollowed by n positive integers. The second line starts with mfollowed 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 32bit signed integer.

输出

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

样例输入

2

4 2 3 4 5

4 6 7 8 9

3 100 200 300

1 150

样例输出

Case 1: 3

Case 2: 0

提示

 

题目来源

Problem Setter: Sohel Hafiz

Special Thanks: Jane Alam Jan

唉...已经不想说话了...后续再补...

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <math.h>
 #include <iostream>
 #include <algorithm>
 #include <climits>
 #include <queue>
 #define ll long long

 using namespace std;

 const int N = ;
 int head[N],total,visit[N];
 int link[N];

 struct nodes
 {
     int e,next;
 } Edge[N];

 void add(int x,int y)
 {
     Edge[total].e = y;
     Edge[total].next = head[x];
     head[x] = total++;
 }

 int dfs(int f)
 {
     for(int i  = head[f]; i != -; i = Edge[i].next)
     {
         int s = Edge[i].e;
         if(visit[s]) continue;
         visit[s] = ;
         if(link[s] == - || dfs(link[s]))
         {
             link[s] = f ;
             return ;
         }
     }
     return ;
 }

 void init()
 {
     total = ;
     memset(head,-,sizeof(head));
     memset(link,-,sizeof(link));
 }

 int main(void)
 {
     int t,a[],b[];
     int i,j,cnt1 = ;
     cin>>t;
     while(t--)
     {
         init();
         int m,n;
         cin>>m;
         for(i = ; i < m; i++)
         {
             scanf("%d",&a[i]);
         }
         cin>>n;
         for(i = ; i < n; i++)
         {
             scanf("%d",&b[i]);
         }
         for(i = ; i < m; i++)
             for(j = ; j < n; j++)
                 if(b[j] % a[i] == )
                 add(i,m+j);
         int cnt;
         for(cnt = ,i = ; i < m+n; i++)
         {
             memset(visit,,sizeof(visit));
             if(dfs(i))
                 cnt++;
         }
         printf("Case %d: %d\n",cnt1++,cnt);
     }

     return ;
 }