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, 
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 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.
尺取直接搞…
然后因为没开longlong wa了一发…
因为看错条件,money差值等于d也会被鄙视．．．所以不能取等号…
都是傻逼错误…药丸，药丸啊…
1/*************************************************************************
2 > File Name: code/cf/#321/B.cpp
3 > Author: 111qqz
4 > Email: rkz2013@126.com
5 > Created Time: 2015年09月25日 星期五 17时21分55秒
6 ************************************************************************/
7
8#include<iostream>
9#include<iomanip>
10#include<cstdio>
11#include<algorithm>
12#include<cmath>
13#include<cstring>
14#include<string>
15#include<map>
16#include<set>
17#include<queue>
18#include<vector>
19#include<stack>
20#include<cctype>
21#define y1 hust111qqz
22#define yn hez111qqz
23#define j1 cute111qqz
24#define ms(a,x) memset(a,x,sizeof(a))
25#define lr dying111qqz
26using namespace std;
27#define For(i, n) for (int i=0;i<int(n);++i)
28typedef long long LL;
29typedef double DB;
30const int inf = 0x3f3f3f3f;
31const int N=1E5+7;
32LL n,d;
33struct Q
34{
35 LL m,f;
36}q[N];
37
38bool cmp(Q a,Q b)
39{
40 if (a.m<b.m) return true;
41 return false;
42}
43
44void solve()
45{
46
47 LL head = 0 ;
48 LL tail = 0 ;
49 LL sum = 0 ;
50 LL ans = -1 ;
51 while (head<n)
52 {
53 while (q[tail].m-q[head].m<d&&tail<n)
54 {
55 sum += q[tail].f;
56// cout<<"sum:"<<sum<<endl;
57 tail++;
58 }
59// cout<<"sum:"<<sum<<endl;
60 ans = max (ans,sum);
61 sum -= q[head].f;
62 head++;
63
64 }
65 cout<<ans<<endl;
66}
67int main()
68{
69 #ifndef ONLINE_JUDGE
70 freopen("in.txt","r",stdin);
71 #endif
72 while(scanf("%lld %lld",&n,&d)!=EOF)
73 {
74 for ( int i = 0; i < n ; i++) cin>>q[i].m>>q[i].f;
75 sort(q,q+n,cmp);
76// for ( int i = 0 ; i < n ; i++) cout<<q[i].m<<" "<<q[i].f<<endl;
77 solve();
78 }
79
80 #ifndef ONLINE_JUDGE
81 fclose(stdin);
82 #endif
83 return 0;
84}