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Senior‘s Gun

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.

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。

1
2 2
2 3
2 2

1

题意：

有n把枪，m只怪物，每把抢都有一个能量值，每个怪物都有一个耗能值，现在用n把枪去打怪物，每把枪只能用一次，怪物只能打一次，用每i把枪打第j只怪物得到能量值为a[i]-b[j],前提a[i]>=b[j]，枪可以不用完，怪物也可以不打完，问最多能得多少的能量值。

解题：枚举用k把枪去打k只怪物，那么每把枪都是最大能量值，每只怪物都是耗能最少的。

<pre name="code" class="cpp">#include<stdio.h>
#include<algorithm>
using namespace std;
const int N = 100005;
bool cmp(int a,int b){
    return a>b;
}
__int64 ans,a[N],b[N];
int main()
{
    
    int T,n,m;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++)
            scanf("%I64d",&a[i]);
        for(int i=0; i<m; i++)
            scanf("%I64d",&b[i]);
        sort(a,a+n,cmp);
        sort(b,b+m);
        int k=0;
        ans=-(1<<29);
        while(k<n&&k<m){
            if(k!=0){
                a[k]+=a[k-1];
                b[k]+=b[k-1];
            }
            if(ans<a[k]-b[k])
                ans=a[k]-b[k];
            k++;
        }
        printf("%I64d\n",ans);
    }
}

HDU 5281 Senior's Gun

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原文地址：http://blog.csdn.net/u010372095/article/details/46853179