CF258D - Little Elephant and Broken Sorting(dp)
阿新 • • 發佈:2021-09-15
題目
有一個長度為\(n\)的排列\(a\),\(m\)次操作交換順序執行,操作\((i,j)\)代表交換\(a_i,a_j\)的值。每次操作等概率發生或不發生。問\(m\)次操作後逆序數對的期望是多少。
題解
經典的轉求貢獻的問題。只要定義好dp狀態就很容易。設\(dp[i][j]\)代表\(a_i>a_j\)的概率。然後轉移方程順帶就出來了。最後答案為
\[\sum\limits_{i<j}{dp[i][j]} \]時間複雜度\(O(n^2)\)
#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; }