Poj 3264 Balanced Lineup【RMQ----ST算法】

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本文介绍了一种使用线段树和RMQ技术优化区间查询的方法，针对特定问题实例——查询区间内最大值与最小值之差，提供了详细的算法实现过程。

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 46109		Accepted: 21658
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 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 from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.
Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

Sample Output

6
3
0

Source

USACO 2007 January Silver

题目大意：给你一个区间为n的序列，有Q个查询，查询区间内最大值和最小值的差。

思路：

1、可以用线段树，也可以用RMQ。

2、RMQ求一遍区间最大值，再求一遍区间最小值，求其差即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<cmath>
#include<iostream>
using namespace std;
int n,q;
int maxn[200005][20];
int minn[200005][20];
void ST()
{
    int len=floor(log10(double(n))/log10(double(2)));
    for(int j=1;j<=len;j++)
    {
        for(int i=1;i<=n+1-(1<<j);i++)
        {
            maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
            minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);
        }
    }
}
int main()
{
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        maxn[i][0]=minn[i][0]=tmp;
    }
    ST();
    for (int i = 1; i <= q; ++i){
            int a,b;
        scanf("%d%d", &a, &b);
        if(a>b)swap(a, b);
        int len= floor(log10(double(b-a+1))/log10(double(2)));
        printf("%d\n",max(maxn[a][len], maxn[b-(1<<len)+1][len])-min(minn[a][len], minn[b-(1<<len)+1][len]));
    }
    return 0;
}