cf:B Kefa and Company

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=94898#problem/B

Kefa and Company

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, 

)
— 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 Input

Input

4 5
75 5
0 100
150 20
75 1

Output

100

Input

5 1000 7
11 32
99 10
46 8
87 54

Output

111

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.

题意：给你一个集合，集合中的每个元素有两个属性，a,b 让你求个子集合，使得集合中的最大m的差不超过d的情况下，b的和的最大值

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 100202
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))

typedef long long LL;

struct node
{
LL a, b;
}p
;

bool cmp (node a, node b)
{
return a.a < b.a;
}

int erfen (int n, LL e);

int main ()
{
LL n, d, sum
;

while (scanf ("%lld %lld", &n, &d) != EOF)
{
for (int i=0; i<n; i++)
scanf ("%lld %lld", &p[i].a, &p[i].b);

sort (p, p+n, cmp);

for (int i=0; i<n; i++)
{
sum[i] = p[i].b;
if (i)
sum[i] += sum[i-1];
}

LL ans = p[0].b;
for (int i=1; i<n; i++)
{
int k = erfen (i, p[i].a-d+1);
ans = max (ans, sum[i]-sum[k]+p[k].b);
}
printf ("%lld\n", ans);
}
return 0;
}

int erfen (int n, LL e)
{
int l = 0, r = n, ans = n, mid;

while (l <= r)
{
mid = (l + r) >> 1;
if (p[mid].a >= e)
{
r = mid - 1;
ans = mid;
}
else l = mid + 1;
}
return ans;
}