poj_3264

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 27473		Accepted: 12898
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

#include<iostream>
#include<cmath>
using namespace std;
#pragma warning(disable : 4996)
#define MAXN 50005
#define min(a, b)   ((a) <= (b) ? (a) : (b))
#define max(a, b)   ((a) >= (b) ? (a) : (b))

int num[MAXN];
int f1[MAXN][100];
int f2[MAXN][100];

//sparse table算法
void st(int n)
{ 
    int i, j, k, m;
    k = (int) (log((double)n) / log(2.0)); 
    for(i = 0; i < n; i++) 
    {
        f1[i][0] = num[i]; //递推的初值
        f2[i][0] = num[i];
    }
    for(j = 1; j <= k; j++)
    { //自底向上递推
        for(i = 0; i + (1 << j) - 1 < n; i++)
        {
            m = i + (1 << (j - 1)); //求出中间的那个值
            f1[i][j] = max(f1[i][j-1], f1[m][j-1]);
            f2[i][j] = min(f2[i][j-1], f2[m][j-1]);
        }
    }
}

//查询i和j之间的最值,注意i是从0开始的
void rmq(int i, int j) 
{ 
    int k = (int)(log(double(j-i+1)) / log(2.0)), t1, t2; //用对2去对数的方法求出k
    t1 = max(f1[i][k], f1[j - (1<<k) + 1][k]);
    t2 = min(f2[i][k], f2[j - (1<<k) + 1][k]);
    printf("%d\n",t1 - t2);
}

int main()
{
    int i, n, q, x, y;
    scanf("%d %d", &n, &q);
    for (i = 0; i < n; ++i)
    {
        scanf("%d", &num[i]);
    }

    st(n); //初始化
    while(q--)
    {
        scanf("%d %d", &x, &y);
        rmq(x - 1, y - 1);
    }
    return 0;
}