CF 258 D. Little Elephant and Broken Sorting

D. Little Elephant and Broken Sorting

链接

题意：

　　长度为n的序列，m次操作，每次交换两个位置，每次操作的概率为$\frac{1}{2}$，求m此操作后逆序对的期望。

分析：

　　f[i][j]表示i>i的概率，每次交换的概率为$\frac{1}{2}$，设交换的位置是x，y，那么$f[i][x]=\frac{f[i][x]+f[i][y]}{2}$，分别是不交换和交换后的概率的和除以2。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<cctype>
 #include<set>
 #include<queue>
 #include<vector>
 #include<map>
 using namespace std;
 typedef long long LL;
 
 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }
 
 const int N = ;
 double f[N][N];
 int a[N];
 
 int main() {
     int n = read(), m = read();
     for (int i = ; i <= n; ++i) a[i] = read();
     for (int i = ; i <= n; ++i)
         for (int j = i + ; j <= n; ++j)
             f[i][j] = a[i] > a[j], f[j][i] = a[j] > a[i];
     while (m --) {
         int x = read(), y = read();
         if (x == y) continue;
         for (int i = ; i <= n; ++i) {
             if (i == x || i == y) continue;
             f[i][x] = f[i][y] = (f[i][x] + f[i][y]) / 2.0;
             f[x][i] = f[y][i] = (f[x][i] + f[y][i]) / 2.0;
         }
         f[x][y] = f[y][x] = 0.5;
     }
     double ans = ;
     for (int i = ; i <= n; ++i)
         for (int j = i + ; j <= n; ++j) ans += f[i][j];
     printf("%.10lf",ans);
     return ;
 }