Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 45243   Accepted: 21240 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

题目链接:http://poj.org/problem? id=3264 Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things s…

题目传送门 Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 64655   Accepted: 30135 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farm…

原题传送门 就是裸RMQ啊.. 求区间最大值和区间最小值,一看就像RMQ,当然线段树貌似也可以. 至于算法嘛.自己学~(好吧,放个传送门...) 然后就是最后把maxsum-minsum就好啦233~ 时间效率:预处理O(N)查找O(1) 是不是很快~ 下面贴代码 #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> using namespace std; int n…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 39046   Accepted: 18291 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

http://poj.org/problem?id=3264 Time Limit: 5000MS     Memory Limit: 65536K Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee w…

本文出自:http://blog.csdn.net/svitter 题意:在1~200,000个数中.取一段区间.然后在区间中找出最大的数和最小的数字.求这两个数字的差. 分析:按区间取值,非常明显使用的线段树. 区间大小取200000 * 4 = 8 * 10 ^5; 进行查询的时候.注意直接推断l, r 与mid的关系就可以.一開始写的时候直接与tree[root].L推断,多余了, 逻辑不对. #include <iostream> #include <stdio.h> #i…

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows fr…

题意:每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John 决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛. 但是为了避免水平悬殊,牛的身高不应该相差太大. John 准备了Q (1 <= Q <= 180,000) 个可能的牛的选择和所有牛的身高 (1 <= 身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别. 注意: 在最大数据上, 输入和输出将占用大部分运…

题目链接:http://poj.org/problem?id=3264 一排牛按1~n标号记录重量,问每个区间最重的和最轻的差值. 线段树维护当前节点下属叶节点的两个最值,查询后作差即可. #include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #include <complex> #includ…