ST算法模板

P3865 【模板】ST 表

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const int N = 1e5 + 10;

int n, m;
int a[N];
int f[N][20];
int mp[N];

void ST_prework() {
    for(int i = 1; i <= n; i++)
        f[i][0] = a[i];
    int t = log(n) / log(2);
    for(int j = 1; j <= t; j++)
        for(int i = 1; i <= n - (1 << j) + 1; i++)
            f[i][j] = max(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
}

int ST_query(int l, int r) {
    int k = mp[r - l + 1];
    return max(f[l][k], f[r - (1 << k) + 1][k]);
}

int main() {
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    for(int i = 0; i <= n; i++)
        mp[i] = log(i) / log(2);
    ST_prework();
    for(int i = 1; i <= m; i++) {
        int l, r;
        scanf("%d%d" ,&l, &r);
        printf("%d\n", ST_query(l, r));
    }
    return 0;
}

 二维ST表，二维矩阵内最值

Cornfields

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const int N = 300;

int n, b, k;
int h[N][N];
int f[N][N][10], F[N][N][10];
int mp[N];

void ST_prework() {
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            f[i][j][0] = h[i][j], F[i][j][0] = h[i][j];
    int t = log2(n);
    for(int k = 1; k <= t; k++)
        for(int i = 1; i <= n - (1 << k) + 1; i++)
            for(int j = 1; j <= n - (1 << k) + 1; j++) {
                f[i][j][k] = max(max(f[i][j][k - 1], f[i + (1 << k - 1)][j][k - 1]),
                                 max(f[i][j + (1 << k - 1)][k - 1], f[i + (1 << k - 1)][j + (1 << k - 1)][k - 1]));
                F[i][j][k] = min(min(F[i][j][k - 1], F[i + (1 << k - 1)][j][k - 1]),
                                 min(F[i][j + (1 << k - 1)][k - 1], F[i + (1 << k - 1)][j + (1 << k - 1)][k - 1]));
            }

}

int ST_query(int x1, int y1, int x2, int y2) {
    int len = mp[b];
    return max(max(f[x1][y1][len], f[x2 - (1 << len) + 1][y1][len]),
               max(f[x1][y2 - (1 << len) + 1][len], f[x2 - (1 << len) + 1][y2 - (1 << len) + 1][len]))
           - (min(min(F[x1][y1][len], F[x2 - (1 << len) + 1][y1][len]),
                  min(F[x1][y2 - (1 << len) + 1][len], F[x2 - (1 << len) + 1][y2 - (1 << len) + 1][len])));
}

int main() {
    cin >> n >> b >> k;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n ; j++)
            scanf("%d", &h[i][j]);
    for(int i = 1; i <= n; i++)
        mp[i] = log2(i);
    ST_prework();
    while(k--) {
        int x, y;
        scanf("%d%d", &x, &y);
        printf("%d\n", ST_query(x, y, x + b - 1, y + b - 1));
    }
    return 0;
}