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每日看点！D. Moscow Gorillas

来源：博客园

D. Moscow Gorillas

In winter, the inhabitants of the Moscow Zoo are very bored, in particular, it concerns gorillas. You decided to entertain them and brought a permutation $p$ of length $n$ to the zoo.

A permutation of length $n$ is an array consisting of $n$ distinct integers from $1$ to $n$ in any order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ occurs twice in the array) and $[1,3,4]$ is also not a permutation ($n=3$, but $4$ is present in the array).

The gorillas had their own permutation $q$ of length $n$. They suggested that you count the number of pairs of integers $l, r$ ($1 \le l \le r \le n$) such that $\operatorname{MEX}([p_l, p_{l+1}, \ldots, p_r])=\operatorname{MEX}([q_l, q_{l+1}, \ldots, q_r])$.


(资料图片)

The $\operatorname{MEX}$ of the sequence is the minimum integer positive number missing from this sequence. For example, $\operatorname{MEX}([1, 3]) = 2$, $\operatorname{MEX}([5]) = 1$, $\operatorname{MEX}([3, 1, 2, 6]) = 4$.

You do not want to risk your health, so you will not dare to refuse the gorillas.

Input

The first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the permutations length.

The second line contains $n$ integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$) — the elements of the permutation $p$.

The third line contains $n$ integers $q_1, q_2, \ldots, q_n$ ($1 \le q_i \le n$) — the elements of the permutation $q$.

Output

Print a single integer — the number of suitable pairs $l$ and $r$.

Examplesinput

31 3 22 1 3

output

2

input

77 3 6 2 1 5 46 7 2 5 3 1 4

output

16

input

61 2 3 4 5 66 5 4 3 2 1

output

11

Note

In the first example, two segments are correct – $[1, 3]$ with $\operatorname{MEX}$ equal to $4$ in both arrays and $[3, 3]$ with $\operatorname{MEX}$ equal to $1$ in both of arrays.

In the second example, for example, the segment $[1, 4]$ is correct, and the segment $[6, 7]$ isn"t correct, because $\operatorname{MEX}(5, 4) \neq \operatorname{MEX}(1, 4)$.

解题思路

比赛的时候想到了通过枚举$\text{mex}$来计算每个$\text{mex}$对答案的贡献是多少，但没写出来。

先考虑$\text{mex} = 1$的情况，首先选取的区间不能包含元素$1$，那么$p$和$q$中能够选取的区间就被分成了$3$个部分，长度分别为$s_1,s_2,s_3$：

那么很明显共有$\dfrac{s_1(s_1 + 1)}{2} + \dfrac{s_2(s_2 + 1)}{2} + \dfrac{s_3(s_3 + 1)}{2}$个区间使得$\text{mex}(p[l \sim r]) = \text{mex}(q[l \sim r]) = 1$。

下面考虑$\text{mex} > 1$的情况。

如果一个排列的区间满足$\text{mex}(p[l \sim r]) = i$，那么区间$[l, r]$一定包含元素$1 \sim i-1$且一定不包含元素$i$。定义$p_x$表示元素$x$在排列中的位置，$\text{minp}[i]$表示前$i$个元素中最小的下标位置，即$\text{minp}[i] = \min\limits_{1 \leq j \leq i} \{ p[j] \}$，同理$\text{maxp}[i] = \max\limits_{1 \leq j \leq i} \{ p[j] \}$。因此如果某个区间的$\text{mex} = i$，那么就要满足$p[i] < \text{minp}[i-1]$或者$p[i] > \text{maxp}[i-1]$，即$p[i]$不能在$\left[ \text{minp}[i-1],\text{maxp}[i-1] \right]$的范围内。

  1. 如果$p_{i} < \text{minp}[i-1]$，那么满足$\text{mex}(p[l \sim r]) = i$的$l$和$r$要满足$p_i < l \leq \text{minp}[i-1]$，$\text{maxp}[i-1] \leq r \leq n$。
  2. 如果$p_{i} > \text{maxp}[i-1]$，那么那么满足$\text{mex}(p[l \sim r]) = i$的$l$和$r$要满足$1 \leq l \leq \text{minp}[i-1]$，$\text{maxp}[i-1] \leq r < p[i]$。
  3. 否则就是$\text{minp}[i-1] \leq p_{i} \leq \text{maxp}[i-1]$，此时没有区间满足$\text{mex}(p[l \sim r]) = i$。

现在是有两个排列$p$和$q$，那么对于同时满足$\text{mex}(p[l \sim r]) = i$和$\text{mex}(q[l \sim r]) = i$的区间可以对各自的$l$和$r$分别通过区间求交的方式得到：

在排列$p$中，$l_p$的取值范围是$[3,5]$（红色部分），$r_p$的取值范围是$[9,13]$（绿色部分）。在$q$中，$l_q$的取值范围是$[1,4]$（蓝色部分），$r_q$的取值范围是$[7,11]$（橙色部分）。

因此能够使得两个排列同时满足$\text{mex}(p[l \sim r]) = i$和$\text{mex}(q[l \sim r]) = i$的$l$要满足$l_p \cap l_q = [3,4]$，$r_p \cap r_q = [9,11]$（记得选取的区间一定使得两个排列都包含元素$1 \sim i-1$），那么满足的区间个数就是$| l_p \cap l_q | \times | r_p \cap r_q | = 2 \times 3 = 6$，即分别是区间$[3, 9]$、$[3,10]$、$[3, 11]$、$[4,9]$、$[4,10]$、$[4,11]$。

AC代码如下：

1 #include  2 using namespace std; 3  4 typedef long long LL; 5  6 const int N = 2e5 + 10; 7  8 int p[N], q[N]; 9 int minp[N], maxp[N], minq[N], maxq[N];10 11 LL get(int a, int b, int c, int d) {    // 区间[a,b]和[c,d]求交集12     return max(0, min(b, d) - max(a, c) + 1);13 }14 15 LL cal(LL x) {  // 计算1+2+...+x16     return x * (x + 1) >> 1;17 }18 19 int main() {20     int n;21     scanf("%d", &n);22     for (int i = 1; i <= n; i++) {23         int x;24         scanf("%d", &x);25         p[x] = i;26     }27     for (int i = 1; i <= n; i++) {28         int x;29         scanf("%d", &x);30         q[x] = i;31     }32     minp[1] = maxp[1] = p[1];33     minq[1] = maxq[1] = q[1];34     for (int i = 2; i <= n; i++) {35         minp[i] = min(minp[i - 1], p[i]);36         maxp[i] = max(maxp[i - 1], p[i]);37         minq[i] = min(minq[i - 1], q[i]);38         maxq[i] = max(maxq[i - 1], q[i]);39     }40     LL ret = 1; // 当两个排列都选区间[1,n]那么就有mex=n+1，这里的1就是指这个情况41     for (int i = 2; i <= n; i++) {  // 考虑mex=2~n的情况42         if (p[i] >= minp[i - 1] && p[i] <= maxp[i - 1] || q[i] >= minq[i - 1] && q[i] <= maxq[i - 1]) continue; // 某个排列无法取得mex=i的区间43         int lp, rp, lq, rq;44         if (p[i] > maxp[i - 1]) lp = 1, rp = p[i] - 1;  // p[i]在区间右边45         else lp = p[i] + 1, rp = n; // p[i]在区间左边46         if (q[i] > maxq[i - 1]) lq = 1, rq = q[i] - 1;  // q[i]在区间右边47         else lq = q[i] + 1, rq = n; //q[i]在区间左边48         ret += get(lp, minp[i - 1], lq, minq[i - 1]) * get(maxp[i - 1], rp, maxq[i - 1], rq);   // 区间求交，然后乘法原理计算答案49     }50     // 最后考虑mex=1的3个区间51     ret += cal(get(1, p[1] - 1, 1, q[1] - 1)) + cal(max(0, abs(p[1] - q[1]) - 1)) + cal(get(p[1] + 1, n, q[1] + 1, n));52     printf("%lld", ret);53     54     return 0;55 }

参考资料

Codeforces Round #852 Editorial：https://codeforces.com/blog/entry/112723

Codeforces Round #852(Div. 2) D(枚举+分类讨论)：https://zhuanlan.zhihu.com/p/605710274

关键词： 同时满足 乘法原理