【题目链接】：click here~~
B. Kefa and Company
time limit per test
memory limit per test
input
output
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
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
Input
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.
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
4 575 5 0 100 150 20 75 1
output
100
input
5 1000 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.
【题目大意】：给你一个集合，集合中的每个元素有两个属性，moner(钱数),factor(友情值) 让你求个子集合，使得集合中的最大money的差不超过d的情况下，factor的和的最大值
【思路】；
我们考虑先按money递增排序，枚举第一个人，然后枚举其他人，依次判断即可，注意此题两个坑点~~
代码：
/*
* Problem: CodeForces 580B
* Running time: 421MS
* Complier: G++
* Author: herongwei
* Create Time: 21:00 2015/10/13 星期二
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
struct node
{
int mon,fac;
bool operator <(const node& t)const{
return mon<t.mon;
}
}st[maxn];
LL sum[maxn]={0};
int main()
{
//freopen("1.txt","r",stdin);
int n,d;cin>>n>>d;
for(int i=1; i<=n; ++i) cin>>st[i].mon>>st[i].fac;
sort(st+1,st+n+1);
for(int i=1; i<=n; ++i) sum[i]=sum[i-1]+st[i].fac;
int l=1,r=1;
LL ans=0;
for(; l<=n; ++l)
{
while(r<=n&&st[r].mon-st[l].mon<d) r++;///写成<=就会WA在第八组数据，没用long long会WA在第九组数据
ans=max(ans,sum[r-1]-sum[l-1]);
}
printf("%lld\n",ans);
return 0;
}
