A - Balanced Lineup

最新推荐文章于 2022-03-26 21:08:54 发布

朱仙令

最新推荐文章于 2022-03-26 21:08:54 发布

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分类专栏： RMQ 文章标签： RMQ

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本文介绍了一种高效解决区间最大最小值之差的问题的方法。通过使用RMQ算法，避免了传统暴力求解带来的效率低下问题。该算法适用于需要频繁查询特定区间内元素的最大值和最小值的应用场景。

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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
题意：一组数字，询问某一区间最大最小值之差

解法：不能用暴力，要超时，用RMQ

<span style="font-size:18px;">#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<math.h>
using namespace std;
int a[50010][17],b[50010][17];
int main()
{
    int n,m,i,j,k;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&k);
        a[i][0]=b[i][0]=k;
    }
    for(k=1;(1<<k)<=n;k++)//RMQ，分别记录区间最大及最小值
        for(i=1;i+(1<<k)-1<=n;i++)
        {
            a[i][k]=max(a[i][k-1],a[i+(1<<(k-1))][k-1]);
            b[i][k]=min(b[i][k-1],b[i+(1<<(k-1))][k-1]);
        }
    while(m--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        k=(int)(log(y-x+1.0)/log(2.0));//公式
        int ma=max(a[x][k],a[y-(1<<k)+1][k]);
        int mi=min(b[x][k],b[y-(1<<k)+1][k]);
        printf("%d\n",ma-mi);
    }
    return 0;
}</span>