poj3264--Balanced Lineup(RMQ求最大最小)-优快云博客

本文介绍如何使用RMQ算法来解决一组奶牛高度差异的问题，通过输入奶牛的数量和潜在的游戏群体，计算每组中最高和最低奶牛的高度差。

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Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 33665		Accepted: 15830
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

用RMQ算法求出最大和最小的值

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[50010] , b[50010][30] , c[50010][30] ;
void RMQ_init1(int n)
{
    int i , j ;
    for(i = 0 ; i < n ; i++)
        b[i][0] = a[i] ;
    for(j = 1 ; (1<<j) <= n ; j++)
    {
        for(i = 0 ; i+(1<<j)-1 < n ; i++)
            b[i][j] = max( b[i][j-1],b[i+(1<<(j-1))][j-1]  ) ;
    }
}
void RMQ_init2(int n)
{
    int i , j ;
    for(i = 0 ; i < n ; i++)
        c[i][0] = a[i] ;
    for(j = 1 ; (1<<j) <= n ; j++)
    {
        for(i = 0 ; i+(1<<j)-1 < n ; i++)
            c[i][j] = min( c[i][j-1],c[ i+( 1<< (j-1) ) ][j-1] ) ;
    }
}
int RMQ1(int l,int r)
{
    int k = 0 ;
    while( 1<<(k+1) <= r-l+1 )
        k++ ;
    return max( b[l][k],b[r-(1<<k)+1][k] ) ;
}
int RMQ2(int l,int r)
{
    int k = 0 ;
    while(1<<(k+1) <= r-l+1 )
        k++ ;
    return min( c[l][k],c[r-(1<<k)+1][k] );
}
int main()
{
    int i , j , n , m , l , r ;
    scanf("%d %d", &n, &m);
    for(i = 0 ; i < n ; i++)
        scanf("%d", &a[i]);
    RMQ_init1(n);
    RMQ_init2(n);
    while(m--)
    {
        scanf("%d %d", &l, &r);
        l-- ; r-- ;
        printf("%d\n", RMQ1(l,r) - RMQ2(l,r) );
    }
    return 0;
}