CF689D Friends and Subsequences 题解 二分答案

题目链接：https://www.luogu.com.cn/problem/CF689D

解题思路：因为

[max_{i=l}^{r} a_i ]

随 (r) 单调递增；

[min_{i=l}^{r} b_i ]

随 (r) 单调递减。

所以我们可以枚举 (l)，二分

[max_{i=l}^{r} a_i = min_{i=l}^{r} b_i ]

对应的 (r) 的左右边界。

区间最值可以用 ST表 维护。

示例代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 200020;
int n, a[maxn], b[maxn], fa[maxn][20], fb[maxn][20];
void init() {
    for (int j = 0; (1<<j) <= n; j ++) {
        for (int i = 1; i+(1<<j)-1 <= n; i ++) {
            if (j == 0) fa[i][j] = a[i], fb[i][j] = b[i];
            else fa[i][j] = max(fa[i][j-1], fa[i+(1<<(j-1))][j-1]), fb[i][j] = min(fb[i][j-1], fb[i+(1<<(j-1))][j-1]);
        }
    }
}
int calmax(int l, int r) {
    int m = r - l + 1;
    int j = log2(m);
    return max(fa[l][j], fa[r-(1<<j)+1][j]);
}
int calmin(int l, int r) {
    int m = r - l + 1;
    int j = log2(m);
    return min(fb[l][j], fb[r-(1<<j)+1][j]);
}
int get_first(int l) {
    int L = l, R = n, res = -1;
    while (L <= R) {
        int mid = (L + R) / 2;
        int v1 = calmax(l, mid), v2 = calmin(l, mid);
        if (v1 >= v2) {
            R = mid - 1;
            if (v1 == v2) res = mid;
        }
        else L = mid + 1;
    }
    return res;
}
int get_last(int l) {
    int L = l, R = n, res = -1;
    while (L <= R) {
        int mid = (L + R) / 2;
        int v1 = calmax(l, mid), v2 = calmin(l, mid);
        if (v1 <= v2) {
            L = mid + 1;
            if (v1 == v2) res = mid;
        }
        else R = mid - 1;
    }
    return res;
}
long long ans;
void solve(int l) {
    int p1 = get_first(l), p2 = get_last(l);
    if (p1 != -1) ans += p2 - p1 + 1;
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++) scanf("%d", a+i);
    for (int i = 1; i <= n; i ++) scanf("%d", b+i);
    init();
    for (int l = 1; l <= n; l ++) solve(l);
    printf("%lld
", ans);
    return 0;
}