Balanced Lineup_poj3264_rmq ST

NO IMAGE

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.

Analysis


強行ST然而這是線段樹例題(/ □ )
ST演算法的核心:
f[i][j]f[i][j]表示從ii往後2j2^j個(ii是第一個)數字區間的最大/最小值
我都想得出來的遞推式

f[i][j]=max(f[i][j−1],f[i 2j−1][j−1])

f[i][j]=\max(f[i][j-1],f[i 2^{j-1}][j-1])
那麼區間i−ji-j的最大值為

max(f[i][i logj−i 12],f[j−2logj−i 12 1][j])

\max(f[i][i log_{2}^{j-i 1}],f[j-2^{log_{2}^{j-i 1}} 1][j])

Code


#include <stdio.h>
#include <math.h>
using namespace std;
int t[50001],maxF[50001][16],minF[50001][16];
int max(int x,int y)
{
return x>y?x:y;
}
int min(int x,int y)
{
return x<y?x:y;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i  )
{
scanf("%d",&t[i]);
maxF[i][0]=minF[i][0]=t[i];
}
for (int j=1;j<=15;j  )
for (int i=1;i<=n&&i (1<<j)-1<=n;i  )
{
maxF[i][j]=max(maxF[i][j-1],maxF[i (1<<(j-1))][j-1]);
minF[i][j]=min(minF[i][j-1],minF[i (1<<(j-1))][j-1]);
}
for (int i=1;i<=m;i  )
{
int x,y;
scanf("%d%d",&x,&y);
int v=floor(log10(y-x 1)/log10(2));
int mx=max(maxF[x][v],maxF[y-(1<<v) 1][v]);
int mn=min(minF[x][v],minF[y-(1<<v) 1][v]);
printf("%d\n",mx-mn);
}
return 0;
}