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

B - Kefa and Company Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 580B Description Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company. Kefa…

B. Kefa and Company Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/580/problem/B Description Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company. Kefa has n friends, each friend wi…

->链接在此<- 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 fri…

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 friend…

http://codeforces.com/problemset/problem/580/B 题意:Kefa有n个朋友,要和这n个朋友中的一些出去,这些朋友有一些钱,并且和Kefa有一定的友谊值,要求这些朋友间的贫富差距(拥有的钱的数量)小于d,要求Kefa邀请出来的朋友的友谊值和最大,并输出这个最大值. #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm&g…

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最大差距小于d并且钱数是最多. 可以用滑动窗口,将m从小到大,s从大到小排列,这时在一个队列里维护队首和队尾,假如队首和队尾的s差距≥d时,就把队尾扔掉队首入队否则就仅队首入队.此时更新一下当前最大值. #include <cstdio> #include <iostream> #include…

题意与分析(CodeForces 580B) \(n\)个人,告诉你\(n\)个人的工资,每个人还有一个权值.现在从这n个人中选出m个人,使得他们的权值之和最大,但是对于选中的人而言,其他被选中的人的工资不能超过他的工资+d. 这题用尺取法可以漂亮的解决.尺取法按照<挑战>一书的定义,是"指对数组保存一对下标(起点和终点),然后根据实际情况交替推进两个端点直到得到答案"的一种方法,因为这种操作"很像是尺蠖(日文中称为尺取虫)爬行的方式"而得名. 具体到这…

排序以后枚举尾部.尺取,头部单调,维护一下就好. 排序O(nlogn),枚举O(n) #include<bits/stdc++.h> using namespace std; typedef long long ll; //#define LOCAL ; struct Node { int m,s; void IN(){ scanf("%d%d",&m,&s); } bool operator < (const Node&rh) const {…

原题链接:http://codeforces.com/contest/580/problem/B 题意: 给你一个集合,集合中的每个元素有两个属性,$m_i,s_i$,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,s的和的最大值. 题解: 先排序,然后对于a[i],直接二分a[i].s+d的位置,然后维护一个前缀和,更新答案即可. 代码: #include<iostream> #include<cstring> #include<algorithm> #inc…