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=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 1005;
double f[N][N];
int a[N];

int main() {
    int n = read(), m = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i <= n; ++i) 
        for (int j = i + 1; 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 = 1; 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 = 0;
    for (int i = 1; i <= n; ++i) 
        for (int j = i + 1; j <= n; ++j) ans += f[i][j];
    printf("%.10lf",ans);
    return 0;
}