poj3264 Balanced Lineup

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.

ST算法模板题。

#include<cstdio>
#include<cstring>
#include<cmath>
int a[50010],mn[50010][20],mx[50010][20];
int min(int x,int y)
{
    return x<y?x:y;
}
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int i,j,k,m,n,p,q,x,y,z,m1,m2;
    scanf("%d%d",&n,&q);
    for (i=1;i<=n;i++)
      scanf("%d",&a[i]),mn[i][0]=mx[i][0]=a[i];
    for (j=1;j<=15;j++)
      for (i=1;i<=n;i++)      
        if (i+(1<<(j-1))>n)
          mn[i][j]=mn[i][j-1],mx[i][j]=mx[i][j-1];
        else
          mn[i][j]=min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]),
          mx[i][j]=max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]);
    for (i=1;i<=q;i++)
    {
        scanf("%d%d",&x,&y);
        k=log(y-x+1)/log(2);
        m1=min(mn[x][k],mn[y-(1<<k)+1][k]);
        m2=max(mx[x][k],mx[y-(1<<k)+1][k]);
        printf("%d\n",m2-m1);
    }
}