Balanced Lineup_poj3264_rmq ST

Description

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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

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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

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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

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强行ST然而这是线段树例题(/ □ )
ST算法的核心：
f[i][j] 表示从 i 往后 2j 个( i 是第一个)数字区间的最大/最小值
我都想得出来的递推式

f[i][j]=max(f[i][j−1],f[i+2j−1][j−1])

那么区间 i−j 的最大值为

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

Code

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#include 
#include 
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 xint 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<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<1][v]);
        int mn=min(minF[x][v],minF[y-(1<1][v]);
        printf("%d\n",mx-mn);
    }
    return 0;
}