zoukankan      html  css  js  c++  java
  • CF div2 321 B

    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.

    读题不清楚真是大问题,然后想二分可以搞,于是读错题的情况下调bug调的想死,然后放弃治疗,换一种方法。

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 #include <stack>
     5 #include <queue>
     6 #include <map>
     7 #include <algorithm>
     8 #include <vector>
     9 #include <cmath>
    10 
    11 using namespace std;
    12 
    13 const int maxn = 1000005;
    14 
    15 typedef long long LL;
    16 
    17 vector<int>G[maxn];
    18 
    19 
    20 struct node
    21 {
    22     int x,y;
    23 }a[maxn];
    24 
    25 
    26 bool cmp(node a,node b)
    27 {
    28     if(a.x == b.x) return a.y<b.y;
    29    return a.x<b.x;
    30 }
    31 
    32 
    33 
    34 int main()
    35 {
    36     int n,m;
    37 
    38     while(scanf("%d%d",&n,&m)!=EOF){
    39     for(int i=1;i<=n;i++){
    40          scanf("%d%d",&a[i].x,&a[i].y);
    41     }
    42 
    43     sort(a+1,a+n+1,cmp);
    44     LL sum = 0;
    45     LL ans = -999999999999;
    46     int l = 1;
    47     for(int i=1;i<=n;i++){
    48         while(a[i].x-a[l].x>=m)
    49             sum -= a[l++].y;
    50             sum += a[i].y;
    51             ans = max(sum,ans);
    52     }
    53 
    54     printf("%lld
    ",ans);
    55 }
    56 
    57     return 0;
    58 }
    View Code
  • 相关阅读:
    Pixar 故事公式
    你想住在中国哪里
    tar.gz方式安装nacos设置使用systemct进行service方式的管理并设置开机自启动
    记一个nginx server_name配置多个时的坑
    linux软链接的创建、修改和删除
    阿里云SLB的健康检查配置
    (转载)bullet安装之——windows下的安装与VS开发
    [译] 找到ndarray中的重复行
    [译] 对dataframe数据按照某列值进行分组,分组后连接字符串
    [译] 如何将列表嵌套列表的情况转化成一维列表
  • 原文地址:https://www.cnblogs.com/lmlyzxiao/p/4834058.html
Copyright © 2011-2022 走看看