uva 11159 Factors and Multiples

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 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 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 (2 from set A and 1 from set B).

Input

The first line of input is an integer T(T<50) that determine the number of test cases. Each case consists of two lines. The first line starts with nfollowed by n integers. The second line starts with m followed by m integers. Both n and m will be in the range [1,100]. All the elements of the two sets will fit in 32 bit signed integer.

Output

For each case, output the case number followed by the answer.

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：

A 和 B 两个集合，从A里面删除一些数 K1 个，从 B里面删除一些数 K2 个，使得 B 中的任何一个元素 都不是 A中 的任意一个数的倍数。

解题思路：

    二部图最大匹配，只要将最大匹配中的全部左边或右边点删掉即可。

解题代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
#include <algorithm>
using namespace std;

const int maxn=110;
int path[maxn][maxn],visited[maxn],link[maxn];
int n,m,a[maxn],b[maxn];

int can(int x){
    for(int i=0;i<m;i++){
        if(visited[i]==-1 && path[x][i]>0){
            visited[i]=1;
            if( link[i]==-1 || can(link[i]) ){
                link[i]=x;
                return 1;
            }
        }
    }
    return 0;
}

void solve(int casen){
    int ans=0;
    for(int i=0;i<n;i++)
    for(int j=0;j<m;j++){
        if(a[i]==0){
            if(b[j]!=0) continue;
            else path[i][j]=1;
        }
        else if(b[j]%a[i]==0){
            path[i][j]=1;
        }
    }
    for(int i=0;i<n;i++){
        memset(visited,-1,sizeof(visited));
        if(can(i)) ans++;
    }
    printf("Case %d: %d\n",casen,ans);
}

int main(){
    //freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    for(int k=1;k<=t;k++){
        memset(path,0,sizeof(path));
        memset(link,-1,sizeof(link));
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d",&a[i]);
        scanf("%d",&m);
        for(int i=0;i<m;i++) scanf("%d",&b[i]);
        solve(k);
    }
    return 0;
}

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uva 11159 Factors and Multiples

原文：http://blog.csdn.net/a1061747415/article/details/23436729