POJ-3264: Balanced Lineup（RMQ）

RMQ

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：输入n和q，然后输入n个数字，q次询问，每次询问输入起点st和终点en，询问这段区间[st,en]里的最大值与最小值的差值是多少，输出结果。

分析： RMQ算法，O(nlogn)的预处理，O(1)地回答每个询问。可以作为模板使用。

• 完整代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[50005],dpmin[50005][16],dpmax[50005][16];
int main()
{
    int i,n,q,st,en,j,d;
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
        {
            dpmin[i][0]=a[i];
            dpmax[i][0]=a[i];
        }
        for(i=1;(1<<i)<= n;i++) //区间长度
            for(j=1;j+(1<<i)-1<=n;j++) //起点
            {
                dpmax[j][i]=max(dpmax[j][i-1],dpmax[j+(1<<i-1)][i-1]);
                dpmin[j][i]=min(dpmin[j][i-1],dpmin[j+(1<<i-1)][i-1]);
            }
        while(q--)
        {
            scanf("%d%d",&st,&en);
            d=0;
            while((1<<d+1)<=en-st+1)  d++;
            printf("%d\n",(max(dpmax[st][d],dpmax[en-(1<<d)+1][d])-min(dpmin[st][d],dpmin[en-(1<<d)+1][d])));
        }
    }
    return 0;
}