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.

Examples
Input

Copy
4 5
75 5
0 100
150 20
75 1
Output

Copy
100
Input

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

Copy
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.

双指针扫一遍即可;

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/ ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
} ll n, d;
struct fr {
ll m, s;
}frd[maxn]; bool cmp(fr x, fr y) {
return x.m < y.m;
} ll sum[maxn]; int main()
{
//ios::sync_with_stdio(0);
rdllt(n); rdllt(d);
for (int i = 1; i <= n; i++)rdllt(frd[i].m), rdllt(frd[i].s);
sort(frd + 1, frd + 1 + n, cmp);
ll maxx = -inf;
int point1 = 1, point2 = 1;
for (int i = 1; i <= n; i++)sum[i] = sum[i - 1] + frd[i].s; while (point2 <= n) {
if (frd[point2].m - frd[point1].m < d) {
maxx = max(maxx, sum[point2] - sum[point1 - 1]);point2++;
}
else {
point1++; } }
cout << maxx << endl;
return 0;
}

CF580B Kefa and Company 尺取法的更多相关文章

  1. [CF580B]Kefa and Company(滑动窗口)

    题目链接:http://codeforces.com/problemset/problem/580/B 某人有n个朋友,这n个朋友有钱数m和关系s两个属性.问如何选择朋友,使得这些朋友之间s最大差距小 ...

  2. Codeforces Round #321 (Div. 2) B. Kefa and Company 二分

    B. Kefa and Company Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/pr ...

  3. CodeForces 580B(尺取法)

    Kefa and Company 题意:Kefa这个人要去吃饭,他要邀请一些朋友一起去,他的每个朋友有两个属性金钱和关系度,要求邀请的人里边任意两个人之间的金钱差的绝对值不大于d:求被邀请的所有朋友的 ...

  4. Unique Snowflakes UVA - 11572 (离散化+尺取法)

    Emily the entrepreneur has a cool business idea: packaging and selling snowflakes. She has devised a ...

  5. 5806 NanoApe Loves Sequence Ⅱ(尺取法)

    传送门 NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K ...

  6. POJ3061 尺取法

    题目大意:从给定序列里找出区间和大于等于S的最小区间的长度. 前阵子在zzuli OJ上见过类似的题,还好当时补题了.尺取法O(n) 的复杂度过掉的.尺取法:从头遍历,如果不满足条件,则将尺子尾 部增 ...

  7. POJ 2739 Sum of Consecutive Prime Numbers(尺取法)

    题目链接: 传送门 Sum of Consecutive Prime Numbers Time Limit: 1000MS     Memory Limit: 65536K Description S ...

  8. CF 701C They Are Everywhere(尺取法)

    题目链接: 传送门 They Are Everywhere time limit per test:2 second     memory limit per test:256 megabytes D ...

  9. CF 321B Kefa and Company(贪心)

    题目链接: 传送门 Kefa and Company time limit per test:2 second     memory limit per test:256 megabytes Desc ...

随机推荐

  1. SRW锁的使用

    SRWLock的目的和关键段相同:对一个资源进行保护,不让其它线程访问它.但是,与关键段不同的是,SRWLock允许我们区分哪些想要读取资源的值 的线程(读取者线程)和想要更新资源的值的线程(写入者线 ...

  2. node与vue结合的前后端分离跨域问题

    第一点:node作为服务端提供数据接口,vue使用axios访问接口, 安装axios npm install axios --save 安装完成后在main.js中增加一下配置: import ax ...

  3. FireMoneky 菜单

    FireMoneky 菜单 TPopup是容器,里边摆放一排button也可. TPopup + ListBox 也可以用弹出窗体实现,form上放一排按钮,功效类似,但是form在fmx下不能半屏显 ...

  4. flask 电影系统(2)

    标签,电影,上映预告数据模型设计 标签数据类型 id:编号 name:标题 movies:电影外键关联 addtime:创建时间 定义电影数据模型 id:编号 title:电影标题 url:电影地址 ...

  5. [Elasticsearch2.x] 多字段搜索 (二) - 最佳字段查询及其调优 <译>

    最佳字段(Best Fields) 假设我们有一个让用户搜索博客文章的网站,就像这两份文档一样: PUT /my_index/my_type/ { "title": "Q ...

  6. 【总结整理】高德LBS开放平台学习

    高德LBS开放平台地址    http://lbs.amap.com/api/javascript-api/guide/create-map/mapstye 概述->示例中心Demo体验-> ...

  7. Linux 下安装redis

    记录一下linux下的安装步骤,还是比较复杂的 1. 下载redis-2.8.19.tar.gz: ftp传到linux01上: 解压: tar –zxvf redis-2.8.19.tar.gz 2 ...

  8. Express响应方法

    下表中响应对象(res)的方法向客户端返回响应,终结请求响应的循环.如果在路由句柄中一个方法也不调用,来自客户端的请求会一直挂起. 方法 描述 res.download() 提示下载文件. res.e ...

  9. 将字符串str1复制为字符串str2的三种方法

    1.自己编写函数,将两个字符串进行复制 #include<iostream> using namespace std; int main(){ char str1[]="I lo ...

  10. 242. Valid Anagram 两个串的最基础版本

    [抄题]: Given two strings s and t, write a function to determine if t is an anagram of s. For example, ...