【LG3322】[SDOI2015]排序

题面

洛谷

题解

交换顺序显然不影响答案，所以每种本质不同的方案就给答案贡献次数的阶乘。

从小往大的交换每次至多\(4\)中决策，复杂度\(O(4^n)\)。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAX_N = 1 << 13;
int N, a[MAX_N];
long long fac[13], ans;
bool check(int x) {
    for (int i = 1; i <= 1 << (N - x); i++)
        if (a[(i - 1) * (1 << x) + 1] + (1 << (x - 1)) !=
            a[(i - 1) * (1 << x) + 1 + (1 << (x - 1))]) return 0;
    return 1;
}
void Swap(int p1, int p2, int k) {
    for (int i = 0; i < k; i++)
        swap(a[p1 + i], a[p2 + i]);
}
void dfs(int x, int p) {
    if (x && !check(x)) return ;
    if (x == N) return (void)(ans += fac[p]);
    dfs(x + 1, p);
    int tmp[5], tot = 0;
    for (int i = 1; i <= 1 << (N - x); i += 2)
        if (a[(i - 1) * (1 << x) + 1] + (1 << x) !=
            a[i * (1 << x) + 1]) {
            if (tot == 4) return ;
            tmp[++tot] = i, tmp[++tot] = i + 1;
        }
    if (!tot) return ;
    for (int i = 1; i <= tot; i++)
        for (int j = i + 1; j <= tot; j++) {
            Swap((1 << x) * (tmp[i] - 1) + 1, (1 << x) * (tmp[j] - 1) + 1, 1 << x);
            dfs(x + 1, p + 1);
            Swap((1 << x) * (tmp[i] - 1) + 1, (1 << x) * (tmp[j] - 1) + 1, 1 << x);
        }
}
int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
#endif
    fac[0] = 1;
    for (int i = 1; i < 13; i++) fac[i] = fac[i - 1] * i;
    scanf("%d", &N);
    for (int i = 1; i <= 1 << N; i++) scanf("%d", a + i);
    dfs(0, 0);
    printf("%lld\n", ans);
    return 0;
}