POJ Topic 3264 Balanced Lineup (RMQ)

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 39046		Accepted: 18291
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'll take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to has 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≤ H Eight ≤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 G Roup.

Input

Line 1:two space-separated integers, Nand Q.
Lines 2.. N+1:line I+1 contains a single integer which is the height of cow I
Lines N+2.. N+ Q+1:two integers Aand B(1≤ A≤ B≤ N), representing the range of cows from ATo BInclusive.

Output

Lines 1.. Q: Each line contains a single integer so is a response to a reply and indicates the difference in height between the Tal Lest and shortest cow in the range.

Sample Input

6 31734251 54) 62 2

Sample Output

630

Source

Usaco January Silver

AC Code

#include <stdio.h> #include <string.h> #include <math.h> #define MAX (b) (a>b?a:b) #define MIN (A, b ) (a>b?b:a) int minv[50050][20],maxv[50050][20];int a[50050];void init (int n) {int i,j,k;for (i=1;i<=n;i++) {maxv[ I][0]=minv[i][0]=a[i];} For (J=1, (1<<j) <=n;j++) {for (k=1;k+ (1<<j) -1<=n;k++) {minv[k][j]=min (minv[k][j-1],minv[k+ (1< < (j-1)][j-1]) Maxv[k][j]=max (maxv[k][j-1],maxv[k+ (1<< (j-1))][j-1]);}}} int Q_max (int l,int r) {int k= (int) (log (double) (r-l+1)/(log (2.0))); return Max (maxv[l][k],maxv[r-(1<<k) +1][k] );} int q_min (int l,int r) {int k= (int) (log (double) (r-l+1)/(log (2.0))); return min (minv[l][k],minv[r-(1<<k) +1][k] );} int main () {int n,m;while (scanf ("%d%d", &n,&m)!=eof) {int i;for (i=1;i<=n;i++) {scanf ("%d", &a[i]);} Init (n), while (m--) {int l,r;scanf ("%d%d", &l,&r);p rintf ("%d\n", Q_max (l,r)-q_min (L,r));}}}

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

POJ Topic 3264 Balanced Lineup (RMQ)