D. Little Elephant and Broken Sorting

分析

題意：

　　長度為n的序列，m次操作，每次交換兩個位置，每次操作的概率為$\frac{1}{2}$，求m此操作後逆序對的期望。

分析：

　　f[i][j]表示i>i的概率，每次交換的概率為$\frac{1}{2}$，設交換的位置是x，y，那麼$f[i][x]=\frac{f[i][x]+f[i][y]}{2}$，分別是不交換和交換後的概率的和除以2。

程式碼：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4
 
 #include<iostream>
 5 #include<cmath>
 6 #include<cctype>
 7 #include<set>
 8 #include<queue>
 9 #include<vector>
10 #include<map>
11 using namespace std;
12 typedef long long LL;
13 
14 inline int read() {
15     int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if
 
(ch=='-')f=-1;
16     for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
17 }
18 
19 const int N = 1005;
20 double f[N][N];
21 int a[N];
22 
23 int main() {
24     int n = read(), m = read();
25     for (int i = 1; i <= n; ++i) a[i] = read();
26     for (int i = 1; i <= n; ++i) 
27
 
         for (int j = i + 1; j <= n; ++j) 
28             f[i][j] = a[i] > a[j], f[j][i] = a[j] > a[i];
29     while (m --) {
30         int x = read(), y = read();
31         if (x == y) continue;
32         for (int i = 1; i <= n; ++i) {
33             if (i == x || i == y) continue;
34             f[i][x] = f[i][y] = (f[i][x] + f[i][y]) / 2.0;
35             f[x][i] = f[y][i] = (f[x][i] + f[y][i]) / 2.0;
36         }
37         f[x][y] = f[y][x] = 0.5;
38     }
39     double ans = 0;
40     for (int i = 1; i <= n; ++i) 
41         for (int j = i + 1; j <= n; ++j) ans += f[i][j];
42     printf("%.10lf",ans);
43     return 0;
44 }