poj 3264 Balanced Lineup

原题：
Balanced Lineup
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 42073 Accepted: 19768
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

6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample Output

6
3
0
Source

USACO 2007 January Silver
大意：
给你两个数，n和q，代表牛的数目和查询的数目。然后给你n个数，和q个查询区间。让你找出在区间当中最大的数和最小的数的差。

#include<iostream>
#include<algorithm>
#include<map>
#include<string>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<vector>
#include<cmath>
#include<stack>
#include<queue>
#include<iomanip>
#include<set>
#include<fstream>
#include <climits>
using namespace std;
//fstream input,output;

int n,q;
int mindp[50010][20];
int maxdp[50010][20];
int cows[50010];

void rmq()
{
    for(int i=1;i<=n;i++)
        mindp[i][0]=maxdp[i][0]=cows[i];
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i+(1<<j)-1<=n;i++)//
        {
            mindp[i][j]=min(mindp[i][j-1],mindp[i+(1<<(j-1))][j-1]);
            maxdp[i][j]=max(maxdp[i][j-1],maxdp[i+(1<<(j-1))][j-1]);
        }
    }
}
int query(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1)
        k++;
    return max(maxdp[l][k],maxdp[r-(1<<k)+1][k]);
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>q;
    for(int i=1;i<=n;i++)
    cin>>cows[i];
    rmq();
    for(int i=0;i<q;i++)
    {
        int a,b;
        cin>>a>>b;
        cout<<query(a,b)<<endl;
    }
//  input.close();
//  output.close();
    return 0;
}

解答：
只是对rmq算法了解过一些，但是没做过题。无意间翻刘汝佳的白书看到了rmq就看了看，然后找到这个入门题目。结果刘汝佳的白书上的代码交上去是错的orz。
裸的rmq，代码编写时注意边界即可。