POJ 3264 Balanced Lineup

Balanced Lineup
Time Limit: 5000MS Memory Limit: 65536K
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.
题目翻译：
平衡的阵容
题目描述：
给你一个有n个数的序列，有Q次询问，每次询问给定一个区间[l,r]，求第l个数到第r个数的区间最大值减去区间最小值。

ST表裸题，我就来扔个板子= =。

#include<algorithm>#include<cstdio>#include<cmath>using namespace std;const int maxn=50000+500;int num[maxn];int st[maxn][30];int st1[maxn][30];int n;void ST(){    for(int i=1;i<=n;i++)        st[i][0]=st1[i][0]=num[i];    int k=(log(n)/log(2));    for(int j=1;j<=k;j++)    {        for(int i=1;i<=n;i++)        {            if(i+(1<<(j-1))<=n)            {                st[i][j]=max(st[i][j-1],st[i+(1<<(j-1))][j-1]);                st1[i][j]=min(st1[i][j-1],st1[i+(1<<(j-1))][j-1]);            }        }    }}int ask_max(int x,int y){    int k=log(y-x+1)/log(2);    return max(st[x][k],st[y-(1<<k)+1][k]);}int ask_min(int x,int y){    int k=log(y-x+1)/log(2);    return min(st1[x][k],st1[y-(1<<k)+1][k]);}int main(){    int q;    scanf("%d%d",&n,&q);    for(int i=1;i<=n;i++)        scanf("%d",&num[i]);    ST();     while(q--)    {        int a,b;        scanf("%d%d",&a,&b);        printf("%d\n",ask_max(a,b)-ask_min(a,b));     }    return 0;}