CF258D - Little Elephant and Broken Sorting(dp)

题目

有一个长度为\(n\)的排列\(a\)，\(m\)次操作交换顺序执行，操作\((i,j)\)代表交换\(a_i,a_j\)的值。每次操作等概率发生或不发生。问\(m\)次操作后逆序数对的期望是多少。

题解

经典的转求贡献的问题。只要定义好dp状态就很容易。设\(dp[i][j]\)代表\(a_i>a_j\)的概率。然后转移方程顺带就出来了。最后答案为

\[\sum\limits_{i<j}{dp[i][j]} \]

时间复杂度\(O(n^2)\)

#include <bits/stdc++.h>

#define endl '\n'
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mp make_pair
#define seteps(N) fixed << setprecision(N) 
typedef long long ll;

using namespace std;
/*-----------------------------------------------------------------*/

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return (__int128)a * b / gcd(a, b);}
#define INF 0x3f3f3f3f

const int N = 3e3 + 10;
const double eps = 1e-5;

double f[N][N];
double g[N][N];
int arr[N];

int main() {
	IOS;
	int n, m;
	cin >> n >> m;
	for(int i = 1; i <= n; i++) {
		cin >> arr[i];
	}
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			f[i][j] = (arr[i] > arr[j]);
		}
	}
	while(m--) {
		int a, b;
		cin >> a >> b;
		for(int i = 1; i <= n; i++) {
			if(i == a || i == b) continue;
			g[i][a] = 0.5 * f[i][b] + 0.5 * f[i][a];
			g[a][i] = 0.5 * f[b][i] + 0.5 * f[a][i];
			g[i][b] = 0.5 * f[i][a] + 0.5 * f[i][b];
			g[b][i] = 0.5 * f[a][i] + 0.5 * f[b][i];
		}
		f[a][b] = 0.5 * f[a][b] + 0.5 * (1 - f[a][b]);
		f[b][a] = 0.5 * f[b][a] + 0.5 * (1 - f[b][a]);
		for(int i = 1; i <= n; i++) {
			if(i == a || i == b) continue;
			f[i][a] = g[i][a];
			f[a][i] = g[a][i];
			f[i][b] = g[i][b];
			f[b][i] = g[b][i];
		}
	}
	double ans = 0;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			if(i >= j) continue;
			ans += f[i][j];
		}
	}
	cout << seteps(10) << ans << endl;
}