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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
Source
题目很水，裸的RMQ自然不必多说，对于每次询问输出区间最大值与最小值绝对值之差，不过要注意输入输出用cin cout会超时，只能用scanf printf，最终速度为1719ms
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define MAXN 50100
using namespace std;
int max(int a,int b)
{
if(a>b) return a;
return b;
}
int min(int a,int b)
{
if(a<b) return a;
return b;
}
int minRMQ[MAXN][18],maxRMQ[MAXN][18];
int getMinRMQ(int l,int r)
{
int pos=(int)(log((r-l+1)*1.0)/log(2.0));
return min(minRMQ[l][pos],minRMQ[r-(1<<pos)+1][pos]);
}
int getMaxRMQ(int l,int r)
{
int pos=(int)(log((r-l+1)*1.0)/log(2.0));
return max(maxRMQ[l][pos],maxRMQ[r-(1<<pos)+1][pos]);
}
int main()
{
int n,m,q;
scanf("%d%d",&n,&q);
for(int i=1;i<=n;i++)
{
scanf("%d",&minRMQ[i][0]);
maxRMQ[i][0]=minRMQ[i][0];
}
m=(int)(log(n*1.0)/log(2.0));
for(int i=1;i<20;i++)
for(int j=1;j<=n;j++)
{
if(j+(1<<i)-1<=n)
{
maxRMQ[j][i]=max(maxRMQ[j][i-1],maxRMQ[j+(1<<(i-1))][i-1]);
minRMQ[j][i]=min(minRMQ[j][i-1],minRMQ[j+(1<<(i-1))][i-1]);
}
}
for(int i=1;i<=q;i++)
{
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",getMaxRMQ(a,b)-getMinRMQ(a,b));
}
return 0;
}
[POJ 3264]Balanced Lineup(ST算法求RMQ)
原文：http://blog.csdn.net/qpswwww/article/details/38826639