AGC014_F - Strange Sorting [递推]

$Description$

Takahashi loves sorting.
He has a permutation ($p_1$,$p_2$,…,$p_N$) of the integers from $1$ through $N$. Now, he will repeat the following operation until the permutation becomes ($1$,$2$,…,$N$):
First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
Then, let $a_1$,$a_2$,…,$a_k$ be the values of the high elements, and $b_1$,$b_2$,…,$b_{N−k}$ be the values of the low elements in the current permutation, in the order they appear in it.
Lastly, rearrange the permutation into ($b_1$,$b_2$,…,$b_{N−k}$,$a_1$,$a_2$,…,$a_k$).
How many operations are necessary until the permutation is sorted?

$Solution$

AGC好难啊，最后还是看了题解做的。。
题解的意思大概是对于一个排列，先不考虑$1$这个数。当经过$T-1$次操作以后排列的第一个元素变成了$f$，可以发现$f>2$，（不然的话会在$T-1$次就已经排序好，或者在$T$次还没有排序好），发现如果此时1在排列中的位置在$f$和$2$之间，答案便是$T$，否则是$T + 1$。另外，有一个很神奇的性质就是这三个数的$cyclic$ $order$是不会变的，即一直是($1$,$2$,$f$)或($2$,$f$,$1$)或($f$,$1$,$2$)，只要在最后判一下这个$cyclic$ $order$就好了。那么这个怎么证明呢。考虑在操作的某时刻有一个数$x$，他是$high$的，并且不在第一个位置，那么必定有一个$y$在$x$之前，且满足$y$是$high$的，$y≤x$，那么经过操作后两个数被放在了一起，那么这两个数会在$y$是$low$的，$x$是$high$的时候分开，又到了最开始情形，一直循环，所以不可能。于是就有了一个结论：当$f$不是这些元素的第一个时，必然是$low$的。那么分类讨论一下$1$,$2$,$f$的位置，就可以证明了。最后问题就相当于从($i+1$,$i+2$,…,$N$)的问题递推到($i$,$i+1$,…,$N$)的问题。
只要一个$for$就好啦：

for (int i = n - 1; i; i--) {
    if (T[i + 1] == 0) {
        if (q[i] > q[i + 1]) {
            T[i] = 1; f[i] = i + 1;
        }
    } else {
        if (Check(q[f[i + 1]], q[i], q[i + 1])) {
            T[i] = T[i + 1]; f[i] = f[i + 1];
        } else {
            T[i] = T[i + 1] + 1; f[i] = i + 1;
        }
    }
}

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 202020;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

int n;
int p[N], q[N];
int f[N], T[N];

inline bool Check(int a, int b, int c) {
    return (a < b && b < c) || (b < c && c < a) || (c < a && a < b);
}

int main(void) {
    freopen("1.in", "r", stdin);
    read(n);
    for (int i = 1; i <= n; i++) {
        read(p[i]); q[p[i]] = i;
    }
    for (int i = n - 1; i; i--) {
        if (T[i + 1] == 0) {
            if (q[i] > q[i + 1]) {
                T[i] = 1; f[i] = i + 1;
            }
        } else {
            if (Check(q[f[i + 1]], q[i], q[i + 1])) {
                T[i] = T[i + 1]; f[i] = f[i + 1];
            } else {
                T[i] = T[i + 1] + 1; f[i] = i + 1;
            }
        }
    }
    cout << T[1] << endl;
}