Kefa and Company

最新推荐文章于 2019-12-23 17:39:01 发布

原创最新推荐文章于 2019-12-23 17:39:01 发布 · 242 阅读

0 ·

CC 4.0 BY-SA版权

OJ练习专栏收录该内容

218 篇文章

订阅专栏

Kefa希望邀请朋友共进晚餐，但要确保没有人在金钱上感到自卑。本篇介绍了如何通过算法帮助Kefa选择最优的朋友组合，以最大化团队的整体友谊度。

L - Kefa and Company

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

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

Hint

In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.

In the second sample test we can take all the friends.

AC代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node {
    int m,s;
} a[100200];
bool cmp(node A,node B) {
    return A.m<B.m;
}
int main() {
    int n,d;
    while(scanf("%d %d",&n,&d)!=EOF) {
        for(int i=0; i<n; i++) {
            scanf("%d %d",&a[i].m,&a[i].s);
        }
        sort(a,a+n,cmp);
        int l=0;
        __int64 max=0,sum=0;
        for(int i=0; i<n; i++) {
            sum+=(__int64)a[i].s;
            while(a[i].m>=a[l].m+d) {
                sum-=(__int64)a[l].s;
                l++;
            }
            if(max<sum)
                max=sum;
        }
        printf("%I64d\n",max);
    }
    return 0;
}