Pku oj 3264 Balanced Lineup（RMQ）

本文主要是介绍Pku oj 3264 Balanced Lineup（RMQ），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 46162		Accepted: 21674
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

线段树也可以过，不过比较慢，ST算法要比线段树快

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;const int maxn = 50005;
int dpmax[maxn][20];
int dpmin[maxn][20];
int a[maxn];
int n,m;void InitRmq()
{for(int i=0;i<n;i++){dpmax[i][0] = a[i];dpmin[i][0] = a[i];}for(int j=1;(1<<j)<=n;j++){for(int i=0;i+(1<<j)-1<n;i++){dpmax[i][j] = max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]);dpmin[i][j] = min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]);}}
}int getMax(int x,int y)
{int k = (int)(log(y-x+1.0)/log(2.0));return max(dpmax[x][k],dpmax[y-(1<<k)+1][k]);
}int getMin(int x,int y)
{int k = (int)(log(y-x+1.0)/log(2.0));return min(dpmin[x][k],dpmin[y-(1<<k)+1][k]);
}int main()
{while(~scanf("%d%d",&n,&m)){for(int i=0;i<n;i++){scanf("%d",&a[i]);}InitRmq();for(int i=0;i<m;i++){int x,y;scanf("%d%d",&x,&y);printf("%d\n",getMax(x-1,y-1)-getMin(x-1,y-1));}}
}

这篇关于Pku oj 3264 Balanced Lineup（RMQ）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！