POJ3264 比较裸的区间最值问题.用线段树或者ST表都可以.此处我们用ST表解决. ST表建表方法采用动态规划的方法, ST[I][J]表示数组从第I位到第 I+2^J-1 位的最值,用二分的思想建立动规方程. 查询方法就不用说了 非常简单. 我这个查询写的稍微有点问题, 理想的ST表查询应该是O(1)的 我写成了O(logn) #include<iostream> #include<cstdio> #include<cstdlib> #include<cst…

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…

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: 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…

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…

  Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 75294   Accepted: 34483 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 J…

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…

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: 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…

/******************************************************* 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…

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 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: 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…

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…

[题目链接] http://poj.org/problem?id=3264 [题目大意] 求区间最大值和最小值的差值 [题解] 线段树维护区间极值即可 [代码] #include <cstdio> #include <algorithm> #include <cstring> #include <climits> using namespace std; const int N=1000010; int T[N*4],C[N*4],n,M,m; 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…

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…

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…

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 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…

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=…

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…

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…

<题目链接> 题目大意: 求给定区间内最大值与最小值之差. 解题分析: 线段树水题,每个节点维护两个值,分别代表该区间的最大和最小值即可. #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define Lson rt<<1,l,mid #define Rson rt<<1|1,mid+1,r #define INF 0x3f…

题意:给出n个数,a1,a2,a3,---,an,再给出q次询问区间al到ar之间的最大值和最小值的差 学习线段树的第一道题目 学习的这一篇 http://www.cnblogs.com/kuangbin/archive/2011/08/14/2137862.html #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #in…

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…

<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…

题意: 求区间max-min st表模板 #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #define N 50010 using namespace std; ],rmax[N][],n,l,r,q; int read() { ,neg=; char j=getchar(); ';j=getchar()) ; ';j=getchar()) ret=ret*+j-…