【学习笔记】ST表

给狂妄自负以适当的绝望，这就是真理

• RMQ问题：
  给定一个长度为N的区间，M个询问，每次询问Li到Ri这段区间元素的最大值/最小值。
  如果暴力找最大值，复杂度是o(n)。但如果查询多次，这个复杂度就很大了。
  解决这个问题的方法是离线ST表和支持在线修改的线段树。

• ST表：一种利用dp求解区间最值的倍增算法。

• 定义：f[i][j]表示i到i+2^j-1这段区间的最大值。

• 预处理：f[i][0]=a[i]。即i到i区间的最大值就是a[i]。

  • 状态转移：将f[i][j]平均分成两段，一段为f[i][j-1]，另一段为f[i+2^(j-1)][j-1]。

  • 两段的长度均为2^j-1。f[i][j]的最大值即这两段的最大值中的最大值。

f[i][j]=max(f[i][j-1],f[i+2^(j-1)][j-1])

• 查询：需要查询的区间为[i,j]，则需要找到两个覆盖这个闭区间的最小幂区间。
  这两个区间可以重复，因为两个区间是否相交对区间最值没有影响。（如下图）
  

模板题：Balanced Lineup

题目描述：

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.

输入：

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.

输出：

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.

样例

6 3
1
7
3
4
2
5
1 5
4 6
2 2

6
3
0

#include <algorithm>
#include <cstdio>
#define ll long long
using namespace std;
const int maxn=1e5;
int stmax[maxn][20],stmin[maxn][20];
int poww[25],logg[maxn];
int n,q;
void init()
{
    poww[0]=1;
    for(int i=1; i<=20; i++)//预处理次方
    {
        poww[i]=poww[i-1]<<1;
    }
    for(int i=2; i<=n; i++)
    {
        logg[i]=logg[i>>1]+1;
    }
    int temp=1;
    for(int j=1; j<=logg[n]; j++)//temp=2^(j-1)
    {
        for(int i=1; i<=n-temp-temp+1; i++)
        {
            stmax[i][j]=max(stmax[i][j-1],stmax[i+temp][j-1]);
            stmin[i][j]=min(stmin[i][j-1],stmin[i+temp][j-1]);
        }
        temp<<=1;
    }
}
inline int query_min(int l,int r)
{
    int len=r-l+1;
    int k=logg[len];
    return min(stmin[l][k],stmin[r-poww[k]+1][k]);
}
inline int query_max(int l,int r)
{
    int len=r-l+1;
    int k=logg[len];
    return max(stmax[l][k],stmax[r-poww[k]+1][k]);
}
int main()
{
    int a;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++){
        scanf("%d",&a);
        stmax[i][0]=stmin[i][0]=a;
    }
    init();
    int l,r;
    while(q--){
        scanf("%d%d",&l,&r);
        printf("%d\n",query_max(l,r)-query_min(l,r));
    }
    return 0;
}

【学习笔记】ST表

原文：https://www.cnblogs.com/smallocean/p/9575850.html