主题链接: PKU:http://poj.org/problem?id=3928 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2492 Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To im…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2492 Ping pong Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank…

Ping pong                                                   Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)   链接:pid=2492">hdu 2492 Problem Description N(3<=N<=20000) ping pong players live along a west-ea…

Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If two players want to compete,…

Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2611    Accepted Submission(s): 973 Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the str…

http://poj.org/problem?id=3928 Ping pong Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2087   Accepted: 798 Description N(3<=N<=20000) ping pong players live along a west-east street(consider the street as a line segment). Each player…

Ping pong Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4874    Accepted Submission(s): 1777 Problem Description N(3<=N<=20000) ping pong players live along a west-east street(consider the st…

UVA 1428 - Ping pong 题目链接 题意:给定一些人,从左到右,每一个人有一个技能值,如今要举办比赛,必须满足位置从左往右3个人.而且技能值从小到大或从大到小,问有几种举办形式 思路:利用树状数组处理出每一个位置左边比它小的个数和右边比他小的个数和.那么左边和右边大就也能计算出来,那么比赛场次为左边小*右边大+左边大*右边小. 代码: #include <cstdio> #include <cstring> const int N = 100005; int t,…

開始用瓜神说的方法撸了一发线段树.早上没事闲的看了一下树状数组的方法,于是又写了一发树状数组 树状数组: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #includ…

对于我这样一名脑残ACMer选手,这道题看了好久好久大概4天,终于知道怎样把它和“树状数组”联系到一块了. 树状数组是什么意思呢?用十个字归纳它:心里有数组,手中有前缀. 为什么要用树状数组?假设你要储存一段数字的前缀和,还要动态修改这些数字.怎么办? 通常的想法就是用数组a[]保存所有的数字,再用数组s[]保存每一位上的前缀和. ;i<=n;i++){ s[i]=s[i-]+a[i]; } 这样有一个弊端,就是当你动态修改a[]数组中的数字的时候,维护s[]数组的代价太高了,最坏可达O(N),…