Balanced Lineup（RMQ）(POJ 3264)

题目：

 RMQ

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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 31734251 54 62 2

Sample Output

630

代码：

/**********************************************************                                                        **     @name:poj 3264                                     **     @author:npufz                                      ***********************************************************//*#include <bits/stdc++.h>*/#include <iostream>#include <cstdlib>#include <cmath>#include <cstdio>using namespace std;int  dp1[50003][16];int  dp2[50003][16];int rmq_st1(int n){    int m=(int )(log(1.0*n)/log(2.0));    for(int j=1;j<=m;j++)    {       int t=n-(1<<j)+1;       for(int i=1;i<=t;i++)       {           dp1[i][j]=min(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);       }   }   return 0;}int rmq_st2(int n){    int m=(int )(log(1.0*n)/log(2.0));    for(int j=1;j<=m;j++)    {       int t=n-(1<<j)+1;       for(int i=1;i<=t;i++)       {           dp2[i][j]=max(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);       }   }   return 0;}int rmq_find1(int l,int r){    int k=(int ) (log(1.0*(r-l+1))/log(2.0));    return min(dp1[l][k],dp1[r-(1<<k)+1][k]);}int rmq_find2(int l,int r){    int k=(int ) (log(1.0*(r-l+1))/log(2.0));    return max(dp2[l][k],dp2[r-(1<<k)+1][k]);}int main(){   int   m,num,left,right,max1,min1;   while(~scanf("%d%d",&num,&m))   {      for(int i=1;i<=num;i++)      {          scanf("%d",&dp1[i][0]);          dp2[i][0]=dp1[i][0];      }       rmq_st1(num);       rmq_st2(num);      for(int i=0;i<m;i++)        {            scanf("%d%d",&left,&right);            min1=rmq_find1(left,right);            max1=rmq_find2(left,right);            printf("%d\n",max1-min1);        }}return 0;}

反思：这是一道很典型的裸题，直接用线段树也可以过，而且省了不少内存，用RMQ_ST也可以过，代码如上