Codeforces#321 (Div. 2) B. Kefa and Company（前缀和，二分查找）

B. Kefa and Company

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ) — 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 mi, si(0 ≤ mi, si ≤ 109) — the amount of money and the friendship factor, respectively.

Output

Print the maximum total friendship factir that can be reached.

Sample test(s)

input

4 5
75 5
0 100
150 20
75 1

output

100

input

5 100
0 7
11 32
99 10
46 8
87 54

output

111

Note

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.

题意：kefa有n个朋友，Kefa 想邀请他的一些朋友去餐厅，每个朋友带两个值mi和si，分别为钱数和友情值，邀请的朋友要求最大的m和最小的m相差不大于d，求能得到的最大的友情值。

思路：按m值递增排序，枚举起点，找到符合要求的终点，计算这个区间的友情值总和，更新答案就好了，为了省时间用前缀和求区间和。注意下数据范围，答案用LL。

#include
#include
#include
#include
using namespace std;

#define LL long long
const LL INF = 1e18;

const int MAXN = 100000 + 1000;

struct node
{
    LL m, s;
    bool operator < (const node& b)const
    {
        return m < b.m;
    }
}num[MAXN];

LL sum[MAXN];
LL M[MAXN];
int n, d;

int main()
{
    while (scanf("%d%d", &n, &d) != EOF)
    {
        for (int i = 1; i <= n; i++)
            scanf("%I64d%I64d", &num[i].m, &num[i].s);
        sort(num + 1, num + n + 1);
        sum[0] = 0;
        M[0] = 0;
        for (int i = 1; i <= n; i++)
            M[i] = num[i].m;
        M[n + 1] = INF;
        for (int i = 1; i <= n; i++)
            sum[i] = sum[i - 1] + num[i].s;
        sum[n + 1] = sum[n];

        LL ans = 0;
        for (int i = 1; i <= n+1; i++)
        {
            int s = i-1;
            int e = ((LL)upper_bound(M + 1, M + n + 2, (LL)M[i] + (LL)d - 1) - (LL)(M + 1)) / sizeof(LL);
            LL temp = sum[e] - sum[s];
            ans = max(ans, temp);
        }
        printf("%I64d\n", ans);
    }
}