[學習筆記]ST表

給狂妄自負以適當的絕望，這就是真理

基本概念

• RMQ問題：
  給定一個長度為N的區間，M個詢問，每次詢問Li到Ri這段區間元素的最大值/最小值。
  如果暴力找最大值，復雜度是\(o(n)\)。但如果查詢多次，這個復雜度就很大了。
  解決這個問題的方法是離線ST表和支持在線修改的線段樹。

• ST表：一種利用dp求解區間最值的倍增算法。

• 定義：\(f[i][j]\)表示\(i\)到\(i+2^{j-1}\)這段區間的最大值。

• 預處理：\(f[i][0]=a[i]\)。即\(i\)到\(i\)區間的最大值就是\(a[i]\)。

  • 狀態轉移：將\(f[i][j]\)平均分成兩段，一段為\(f[i][j-1]\)，另一段為\(f[i+2^{j-1}][j-1]\)。

  • 兩段的長度均為\(2^{j-1}\)。\(f[i][j]\)的最大值即這兩段的最大值中的最大值。

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

• 查詢：需要查詢的區間為\([i,j]\)，則需要找到兩個覆蓋這個閉區間的最小冪區間。
  這兩個區間可以重復，因為兩個區間是否相交對區間最值沒有影響。（如下圖）
  

模板題：Balanced Lineup

• 題目描述：

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.

• 輸入：

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.

• 輸出：

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.

• 樣例

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

6
3
0

代碼及模板

#include <bits/stdc++.h>

#define ll long long
using namespace std;
const int maxn = 1e5 + 7;
int stmax[maxn][20], stmin[maxn][20];
int poww[25], logg[maxn];
int n, q;

void init() {
    poww[0] = 1;//預處理次方
    for (int i = 1; i <= 20; i++) poww[i] = poww[i - 1] << 1;
    for (int i = 2; i <= n; i++) logg[i] = logg[i >> 1] + 1;
    int temp = 1;//temp=2^(j-1)
    for (int j = 1; j <= logg[n]; j++, temp <<= 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]);
        }
    }
}

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;
}

LCA+ST的模板

用dfs序表示2*n的數組，數組的值是深度。fir表示第一次遍歷到這個點的dfs序，cur表示這個dfs序對應的節點。lca就是l到r區間深度最小的節點。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=5e5+10;//點的個數
int cur[maxn<<1];// cur 當前dfs序的點
int fir[maxn];//fir 第一次遍歷的 dfs序
int rmq[maxn<<1];//rmq 深度
struct Edge{
    int to,next;
}e[maxn<<1];
int head[maxn],tol,cnt;
struct St{
    int stm[maxn<<1][21];//stm dfs序列上深度最小的點
    int logg[maxn<<1],poww[21];
    void init(int n) {
        poww[0]=1;logg[0]=-1;
        for(int i=1; i<=20; i++)
        {
            poww[i]=poww[i-1]<<1;
        }
        for (int i = 1; i <= n; i++) {
            logg[i]=logg[i>>1]+1;
            stm[i][0]=i;
        }
        int temp=1;
        for (int j = 1; j <=logg[n]; ++j,temp<<=1) {
            for (int i = 1; i <=n-temp-temp+1; ++i) {
                stm[i][j]=rmq[stm[i][j-1]]<rmq[stm[i+temp][j-1]]?stm[i][j-1]:stm[i+temp][j-1];
            }
        }
    }
    int Query(int a,int b){
        if(a>b) swap(a,b);
        int k=logg[b-a+1];
        return rmq[stm[a][k]]<=rmq[stm[b-poww[k]+1][k]]?stm[a][k]:stm[b-poww[k]+1][k];
    }
}st;
void add(int u,int v){
    e[++tol].to=v;
    e[tol].next=head[u];
    head[u]=tol;
}
void dfs(int u,int pre,int dep){
    cur[++cnt]=u;
    rmq[cnt]=dep;
    fir[u]=cnt;
    for(int i=head[u];i;i=e[i].next){
        int v=e[i].to;
        if(v==pre) continue;
        dfs(v,u,dep+1);
        cur[++cnt]=u;
        rmq[cnt]=dep;
    }
}
void cal(int roof,int n){
    dfs(roof,roof,0);
    st.init(n);
}
int query_lca(int a,int b){
    return cur[st.Query(fir[a],fir[b])];
}
int main(){
     int n,m,s,u,v;
     scanf("%d%d%d",&n,&m,&s);
    for (int i = 1; i < n; ++i) {
        scanf("%d%d",&u,&v);
        add(u,v);
        add(v,u);
    }
    cal(s,(n*2)-1);
    while (m--){
        scanf("%d%d",&u,&v);
        printf("%d\n",query_lca(u,v));
    }
}