ST表

ST表（Sparse Table，稀疏表）是一种简单的数据结构，主要用来解决RMQ（Range Maximum/Minimum Query，区间最大/最小值查询）问题。

//进行预处理，计算区间最大值
int f[MAXN][21]; // 第二维的大小根据数据范围决定，不小于log(MAXN)
for (int i = 1; i <= n; ++i)
    f[i][0] = read(); // 读入数据
for (int i = 1; i <= 20; ++i)
    for (int j = 1; j + (1 << i) - 1 <= n; ++j)
        f[j][i] = max(f[j][i - 1], f[j + (1 << (i - 1))][i - 1]);

//预处理log计算
for (int i = 2; i <= n; ++i)
    Log2[i] = Log2[i / 2] + 1;

//查询
for (int i = 0; i < m; ++i)
{
    int l = read(), r = read();
    int s = Log2[r - l + 1];
    printf("%d\n", max(f[l][s], f[r - (1 << s) + 1][s]));
}

例题：
（洛谷P2880 [USACO07JAN]平衡的阵容Balanced Lineup）

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 ≤ 180,000) 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.

#include<bits/stdc++.h>
using namespace std;
int Log[50000+2],Max[50000+2][20],Min[50000+2][20];
int main() {
    int n,m;
    cin >> n >> m;
    for(int i = 1;i <= n; ++ i) {
        cin >> Max[i][0];
        Min[i][0] = Max[i][0];
    }
    for(int i = 2;i <= n; ++ i) {
        Log[i] = Log[i / 2] + 1;
    }
    for(int i = 1;i < 20; ++ i) {
        for(int j = 1;j + (1 << i) - 1 <= n; ++ j) {
            Min[j][i] = min(Min[j][i - 1], Min[j + (1 << (i - 1))][i - 1]);
            Max[j][i] = max(Max[j][i - 1], Max[j + (1 << (i - 1))][i - 1]);
        }
    }
    for(int i = 0;i < m; ++ i) {
        int l,r;
        cin >> l >> r;
        int s = Log[r - l + 1];
        int ma = max(Max[l][s], Max[r - (1 << s) + 1][s]);
        int mi = min(Min[l][s], Min[r - (1 << s) + 1][s]);
        printf("%d\n", ma - mi);
    }
    return 0;
}