【LG3322】[SDOI2015]排序

【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;
}