题目链接:http://poj.org/problem?id=3264 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 wil…

http://poj.org/problem?id=3264 题目大意: 给定N个数,还有Q个询问,求每个询问中给定的区间[a,b]中最大值和最小值之差. 思路: 依旧是线段树水题~ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int MAXN=50000+10; const int MAXM=MAXN<<2; const int INF=…

题目链接: Poj 3264 Balanced Lineup 题目描述: 给出一个n个数的序列,有q个查询,每次查询区间[l, r]内的最大值与最小值的绝对值. 解题思路: 很模板的RMQ模板题,在这里总结一下RMQ:RMQ(Range Minimum/Maximum Query) 即区间最值查询,是指这样一个问题:对于长度为n的数列A,回答若干询问RMQ(A,i,j)(i,j<=n),返回数列A中下标在i,j之间的最小/大值. RMQ有三种求法:1:直接遍历查找,炒鸡暴力: 2:线段树也可以解…

传送门:http://poj.org/problem?id=3264 Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 66241   Accepted: 30833 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in…

<题目链接> 题目大意: 给定一段序列,进行q次询问,输出每次询问区间的最大值与最小值之差. 解题分析: RMQ模板题,用ST表求解,ST表用了倍增的原理. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long ll; ; int n,q; ll maxsum[M][],mi…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 43168   Accepted: 20276 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: Balanced Lineup(poj 3264) 链接: http://poj.org/problem?id=3264 题意: 给个数列,查询一段区间的最大值与最小值的差 算法: RMQ ********************************************************/ #include<cstdio> #include<cstrin…

点我看题目 题意 :N头奶牛,Q次询问,然后给你每一头奶牛的身高,每一次询问都给你两个数,x y,代表着从x位置上的奶牛到y位置上的奶牛身高最高的和最矮的相差多少. 思路 : 刚好符合RMQ的那个求区间最大最小值,所以用RMQ还是很方便的.就是一个RMQ的模板题,基本上书上网上都有. RMQ基础知识 RMQ算法举例 #include <stdio.h> #include <string.h> #include <math.h> #include <iostream…

题目地址:http://poj.org/problem?id=3264 Sample Input 6 3 1 7 3 4 2 5 1 5 4 6 2 2 Sample Output 6 3 0分析:标准的模板题,可以用线段树写,但用RMQ-ST来写代码比较短.每次输出区间[L, R]内最大值和最小值的差是多少.注意一个地方,代码里面用到了log2()函数,但是我用包含<math.h>和<cmath>头文件的代码以C++的方式提交到POJ反馈是编译错误.改成g++提交才AC了.(注意…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 32820   Accepted: 15447 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…

RMQ (Range Minimum/Maximum Query)问题是指:对于长度为n的数列A,回答若干询问RMQ(A,i,j)(i,j<=n),返回数列A中下标在i,j里的最小(大)值,也就是说.RMQ问题是指求区间最值的问题. id=10244" target="_blank" style="color:blue; text-decoration:none">Balanced Lineup Time Limit: 5000MS   Mem…

Balanced Lineup 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 simple, he will take a…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 53703   Accepted: 25237 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…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 68466   Accepted: 31752 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…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status 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 organi…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 34306   Accepted: 16137 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…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 34306   Accepted: 16137 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…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 42489   Accepted: 20000 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…

这是线段树的一个模板题,给出一串数字,然后询问区间的最大最小值. 这个其实很好办,只需把线段树的节点给出两个权值,一个是区间的最小值,一个是区间的最大值,初始化为负无穷和正无穷,然后通过不断地输入节点,不断维护,最好每次询问维护一个询问区间的最大值和最小值,最后相减即可.其实就相当于,线段树找区间的最大值和最小值. #include<iostream> #include<stdio.h> #include<algorithm> #include<string.h&…

Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 62103 Accepted: 29005 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 John de…

Balanced Lineup 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…

RMQ支持操作: Query(L, R):  计算Min{a[L],a[L+1], a[R]}. 预处理时间是O(nlogn), 查询只需 O(1). RMQ问题 用于求给定区间内的最大值/最小值问题..询问的次数多的时候 好用.. 这个题目我至少得开数组开到 80000才能过,不知道为什么..刚开始还写错了,贡献了好多RE和WA.. 题目:http://poj.org/problem?id=3264 题意:给n个数,q次询问,求最值的差.. #include <iostream> #incl…

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 simple, he will take a contiguous rang…

题目链接: http://poj.org/problem?id=3264 思路分析: 典型的区间统计问题,要求求出某段区间中的极值,可以使用线段树求解. 在线段树结点中存储区间中的最小值与最大值:查询时使用线段树的查询 方法并稍加修改即可进行查询区间中最大与最小值的功能. 代码(线段树解法): #include <limits> #include <cstdio> #include <iostream> using namespace std; ; + ; struct…

链接:http://poj.org/problem?id=3264 题意:给n个数,求一段区间L,R的最大值 - 最小值,Q次询问 思路:ST表模板,预处理区间最值,O(1)复杂度询问 AC代码: #include<iostream> #include<vector> #include<cstdlib> #include<cstdio> #include<algorithm> #include<cmath> #include<s…

题目:http://poj.org/problem?id=3264 给定一段区间,求其中最大值与最小值的差. #include <stdio.h> #include <algorithm> #include <math.h> ][], dpMax[][]; int RMinQ(int l, int r) { )/log(2.0); <<k)+][k]); } int RMaxQ(int l, int r) { )/log(2.0); <<k)+]…

一段区间的最值问题,用线段树或RMQ皆可.两种代码都贴上:又是空间换时间.. RMQ 解法:(8168KB 1625ms) #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <cstdlib> using namespace std; #define N 50003 ],dmax…

这个题目是一个典型的RMQ问题,给定一个整数序列,1~N,然后进行Q次询问,每次给定两个整数A,B,(1<=A<=B<=N),求给定的范围内,最大和最小值之差. 解法一:这个是最初的解法,时间上可能会超时,下面还有改进算法.4969ms #include<stdio.h> #include<stdlib.h> #define INF 1000010 #define max(a,b) ((a)>(b)?(a):(b)) #define min(a,b) ((a…

题意:有n头牛,编号从1到n,每头牛的身高已知.现有q次询问,每次询问给出a,b两个数.要求给出编号在a与b之间牛身高的最大值与最小值之差. 思路:标准的RMQ问题. RMQ问题是求给定区间内的最值问题.当询问量巨大时,最朴素算法必然超时.解决RMQ比较优秀的算法有ST算法.其预处理时间复杂度为O(nlogn),询问的时间复杂度为O(1). ST的思想如下: 假设num数组中的数据从第0位开始存储. 用两个二维数组tmax,tmin分别求区间最大与最小值.ST的关键是数组区间的分割.tmax和t…

题意:求区间最大值-最小值. 分析: 1.线段树 #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<st…