POJ3264
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 2Sample Output
6
3
0这道题是区间查询问题，牵扯到多次查询，需要对区间最值进行预处理。直接套ST算法即可。代码如下：
详解：RMQ
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=5e4+10;
int n,q,a[maxn],dmax[maxn][20],dmin[maxn][20];//ST算法
void RMQ(){for(int i=1;i<=n;i++){dmax[i][0]=a[i];dmin[i][0]=a[i];}for(int j=1;(1<<j)<=n;j++){for(int i=1;i+(1<<j)-1<=n;i++){dmax[i][j]=max(dmax[i][j-1],dmax[i+(1<<(j-1))][j-1]);dmin[i][j]=min(dmin[i][j-1],dmin[i+(1<<(j-1))][j-1]);}}
}int main(){scanf("%d%d",&n,&q);for(int i=1;i<=n;i++){scanf("%d",&a[i]);}RMQ();int l,r;while(q--){scanf("%d%d",&l,&r);int k=(int)(log(r-l+1.0)/log(2.0));int mx=max(dmax[l][k],dmax[r-(1<<k)+1][k]);int mn=min(dmin[l][k],dmin[r-(1<<k)+1][k]);printf("%d\n",mx-mn);}return 0;
}