Senior‘s Gun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 842    Accepted Submission(s): 309

Problem Description

Xuejiejie is a beautiful and charming sharpshooter.

She often carries n guns, and every gun has an attack power a[i].

One day, Xuejiejie goes outside and comes across m monsters, and every monster has a defensive power b[j].

Xuejiejie can use the gun i to kill the monster j, which satisfies b[j]≤a[i], and then she will get a[i]−b[j] bonus .

Remember that every gun can be used to kill at most one monster, and obviously every monster can be killed at most once.

Xuejiejie wants to gain most of the bonus. It‘s no need for her to kill all monsters.

Input

In the first line there is an integer T, indicates the number of test cases.

In each case:

The first line contains two integers n, m.

The second line contains n integers, which means every gun‘s attack power.

The third line contains m integers, which mean every monster‘s defensive power.

1≤n,m≤100000, −109≤a[i],b[j]≤109。

Output

For each test case, output one integer which means the maximum of the bonus Xuejiejie could gain.

Sample Input

1
2 2
2 3
2 2

Sample Output

1

Source

BestCoder Round #47 ($)

/**
    题意：xuejiejie有n把枪，每把枪的威慑力是mmap[i]，现在有m个master,每个master
          的攻击性是temp[i] ，现在xuejiejie得到的bonus值为mmap[i] - temp[i]
          现在要求bonus的值最大化
    做法：暴力
**/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxn 100000 + 10
#define INF 0x7fffffff
long long  mmap[maxn];
long long  temp[maxn];
int main()
{
//#ifndef ONLINE_JUDGE
//    freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d %d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&mmap[i]);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%lld",&temp[i]);
        }
        sort(mmap,mmap+n);
        sort(temp,temp+m);
        long long sum = 0;
        long long ans = 0;
        int p = 0;
        for(int i=n-1;i>=0;i--)
        {
            if(p>=m) break;
            if(mmap[i] >= temp[p])
            {
               ans += mmap[i] - temp[p];
               sum = max(sum,ans);
               p++;
            }
            else
            {
                p++;
                i++;
            }
        }
        printf("%lld\n",sum);
    }
    return 0;
}