自拉比赛B题（区间更新）

最新推荐文章于 2024-02-18 21:42:40 发布

原创最新推荐文章于 2024-02-18 21:42:40 发布 · 557 阅读

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本文介绍了一个简单的区间更新问题，通过排序和滑动窗口的方法来解决Kefa如何邀请朋友以达到最大友谊值的问题。该问题涉及输入输出、排序及数组操作等基本算法知识。

Description

Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.

Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!

Input

The first line of the input contains two space-separated integers, n and d (1 ≤ n ≤ 10⁵, $\text{[math]}$ ) — the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.

Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type m_i, s_i (0 ≤ m_i, s_i ≤ 10⁹) — the amount of money and the friendship factor, respectively.

Output

Print the maximum total friendship factir that can be reached.

Sample Input

Input

Output

Input

Output

题解：区间更新思路。题比较水。

直接看代码吧：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
const int maxn=100010;
struct py//定义一个结构体数组，赋予其中两个含义：金钱和人品
{
    long long qian;
    long long renpin;
} flog[maxn];
bool cmp(py x,py y)//定义升序排列金钱函数
{
    return x.qian<y.qian;
}
int main()
{
    int n,m;
    long long s[maxn];//存储可加人品区间的总值
    long long maxx=0;//最大的人品
    while(~scanf("%d%d",&n,&m))
    {
        for(int i =1; i<=n; i++)
        {
            scanf("%lld",&flog[i].qian);//赋值金钱
            scanf("%lld",&flog[i].renpin);//赋值人品
        }
        sort(flog+1,flog+n+1,cmp);//调用cmp进行金钱排序
        memset(s,0,sizeof(s));
        int k=1;

        for(int j=1; j<=n; j++)
        {
            s[j]=s[j-1]+flog[j].renpin;//求可加区间人品总值
            maxx=max(maxx,flog[j].renpin);//求出最大的人品
            while(flog[j].qian-flog[k].qian>=m)//若不符合金钱差，更新可加区间第一个值
            {
                k++;
            }
            maxx=max(maxx,s[j]-s[k-1]);//求出最终的最大人品值

        }

      printf("%lld",maxx);//输出结果
    }
    return 0;

}