poj--3264 Balanced Lineup（裸的RMQ）

Posted 2020-08-15 Wally的博客

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

6 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0
题意：给你一个数列，不停地询问你区间【L，R】的最大值和最小值之差。
思路：由于数据特别大，所以暴力的话一定会超时的。所以考虑到用 RMQ ，经过简单的学习就能把这个题目给A了。
RMQ博客：http://blog.csdn.net/liang5630/article/details/7917702

AC代码：

 1 #include <iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 int a[200005],b[200005][40],c[200005][40];
 5 int main()
 6 {
 7     int n,m,l,r;
 8     while(~scanf("%d%d",&n,&m))
 9     {
10         for(int i=1; i<=n; i++)
11         {
12             scanf("%d",&a[i]);
13             b[i][0]=c[i][0]=a[i];
14         }
15         for(int j=1; (1<<j)<=n; j++)
16             for(int i=1; i+(1<<(j-1))<=n; i++)
17             {
18                 b[i][j]=max(b[i][j-1],b[i+(1<<(j-1))][j-1]);
19                 c[i][j]=min(c[i][j-1],c[i+(1<<(j-1))][j-1]);
20                 //printf("%d %d %d %d\\n",i,j,b[i][j],c[i][j]);
21             }
22         for(int i=0; i<m; i++)
23         {
24             scanf("%d%d",&l,&r);
25             int k=0;
26             while((1<<k)<=r-l+1)
27                 k++;
28             printf("%d\\n",max(b[l][k-1],b[r-(1<<(k-1))+1][k-1])-min(c[l][k-1],c[r-(1<<(k-1))+1][k-1]));
29         }
30     }
31     return 0;
32 }

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