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 = 20000;
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 != -1; i = Edge[i].next)
    {
        int s = Edge[i].e; 
        if(visit[s]) continue;
        visit[s] = 1;
        if(link[s] == -1 || dfs(link[s]))
        {
            link[s] = f ;
            return 1;
        }
    }
    return 0;
}

void init()
{
    total = 0;
    memset(head,-1,sizeof(head));
    memset(link,-1,sizeof(link));
}

int main(void)
{
    int t,a[105],b[105];
    int i,j,cnt1 = 1;
    cin>>t;
    while(t--)
    {
        init();
        int m,n;
        cin>>m;
        for(i = 0; i < m; i++)
        {
            scanf("%d",&a[i]);
        }
        cin>>n;
        for(i = 0; i < n; i++)
        {
            scanf("%d",&b[i]);
        }
        for(i = 0; i < m; i++)
            for(j = 0; j < n; j++)
                if(b[j] % a[i] == 0)
                add(i,m+j);
        int cnt;
        for(cnt = 0,i = 0; i < m+n; i++)
        {
            memset(visit,0,sizeof(visit));
            if(dfs(i))
                cnt++;
        }
        printf("Case %d: %d
",cnt1++,cnt);
    }

    return 0;
}