CF6E Exposition 题解 ST表+倍增

题目大意：

求所有极差不超过 \(k\) 的最长连续子序列。

解题思路：

先开一个 ST 表方便求解区间最大值和区间最小值。

然后基于倍增思想（详见 cal 函数）求极差不超过 \(k\) 的最长连续子序列。

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int n, k, mx[maxn][18], mi[maxn][18], a[maxn], res[maxn];

int cal(int p) {
    int len = 1, x = a[p], y = a[p];
    for (int i = 17; i >= 0; i--)
        if (p + (1<<i) <= n && max(x, mx[p+1][i]) - min(y, mi[p+1][i]) <= k)
            len += 1<<i,
            x = max(x, mx[p+1][i]),
            y = min(y, mi[p+1][i]),
            p += 1<<i;
    return len;
}

int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)
        scanf("%d", a+i), mx[i][0] = mi[i][0] = a[i];
    for (int j = 1; j < 18; j++)
        for (int i = 1; i+(1<<j)-1 <= n; i++)
            mx[i][j] = max(mx[i][j-1], mx[i+(1<<j-1)][j-1]),
            mi[i][j] = min(mi[i][j-1], mi[i+(1<<j-1)][j-1]);
    int x = 0, cnt = 0;
    for (int i = 1; i <= n; i++) {
        res[i] = cal(i);
        if (res[i] > x) x = res[i], cnt = 1;
        else if (res[i] == x) cnt++;
    }
    printf("%d %d\n", x, cnt);
    for (int i = 1; i <= n; i++)
        if (res[i] == x)
            printf("%d %d\n", i, i+x-1);
    return 0;
}