[BZOJ5449] Sequence

Title link: Sequence

Description

Given an arrangement x of \ (1 \) ~ \ (n \) , you can flip \ (x_1, x_2, ..., x_i \) each time .
You need to ask for the minimum number of operations to change the sequence into ascending order.
Multiple sets of data.
Data range \ (T = 5,1 \ le n \ le 25 \)
Time limit \ (10 ​​\ s \)

Solution

The data range is so small, we consider using \ (IDA * \) to optimize burst search.
Define the valuation function \ (h () = \ sum_ {i = 2} ^ {n} [abs (x_i-x_ {i-1}) ≠ 1] \)
Considering each flip, you can make at most \ (abs ≠ 1 The logarithm of \) is reduced by one, so you can use \ (h () \) to estimate.
Just search for it directly.
Complexity \ (O (Can pass) \)

Code

// Author: wlzhouzhuan
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define rint register int
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

int a[26], flag, n, maxd;
int h() {
  int cnt = 0;
  for (rint i = 2; i <= n; i++) if (abs(a[i] - a[i - 1]) != 1) {
    cnt++;
  }
  return cnt;
}
bool check() {
  for (rint i = 2; i <= n; i++) {
    if (a[i - 1] > a[i]) {
      return 0;
    }
  }
  return 1;
}
void dfs(int x, int last) { // x表示操作次数，last表示上一次翻转的位置 
  if (flag) return ;
  if (x == maxd) {
    if (check()) flag = 1;
    return ;
  }
  if (x + h() > maxd) {
    return ;
  } 
  for (rint i = 2; i <= n; i++) if (i != last) {
    reverse(a + 1, a + i + 1);
    dfs(x + 1, i);
    reverse(a + 1, a + i + 1); 
  }
}
int main() {
  int T = read();
  while (T--) {
    n = read();
    for (int i = 1; i <= n; i++) {
      a[i] = read();
    }
    flag = 0;
    for (maxd = 0; maxd <= 30; maxd++) {
      dfs(0, 0);
      if (flag) break;
    }
    printf("%d\n", maxd);
  }
  return 0;
}