POJ - 3264 Balanced Lineup 线段树解RMQ】的更多相关文章

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…
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: 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 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 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…
Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 34140   Accepted: 16044 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: 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…
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: 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…