Codeforces 258d little elephant and broken sorting (expected)

Cf258d little elephant and broken sorting

Question Translation

There is an \ (1 \ SIM n \) arrangement, the \ (M \) operation is performed, the operation is exchange \ (a, B \). Each operation has a probability of \ (50 \ %.

Evaluate the number of expected Reverse Order pairs after performing the \ (M \) operation.

\ (N, m \ le 1000 \)

Input/output format:

The first line contains two integers \ (n \) and \ (M \) \ (1 \ Leq n, m \ Leq 1000, n> 1 )\) -The permutation size and the number of moves. the second line contains \ (n \) distinct integers, not exceeding \ (n \)-the initial permutation. next \ (M \) lines each contain two integers: The \ (I \)-th line contains integers \ (A _ {I }\) and \ (B _ {I} \) \ (1 \ Leq A _ {I}, B _ {I} \ Leq N, A _ {I} \ neq B _ {I}) \)-the positions of elements that were changed during the \ (I \)-th move.

Output Format:

In the only line print a single real number-the answer to the problem. the answer will be considered correct if its relative or absolute error does not exceed \ (10 ^ {-6 }\).

Input and Output sample input sample #1:

2 11 21 2

Output sample #1:

0.500000000

Input example #2:

4 31 3 2 41 22 31 4

Output sample #2:

3.000000000

Ideas

This question is true. -- Mercury

I would like to thank the mercury for his unexpected status design.

Define \ (f (I, j) \) as the probability that the ratio of numbers on the position \ (I \) is greater than that on the position \ (J. Assuming that each exchange is successful \ (100 \ % \), you may set the subscript of the number of this exchange as \ (a, B \), then any \ (f (I, A), F (I, B) \) will be exchanged, \ (f (a, I), F (B, I) \) will also be exchanged. However, the probability of the current switch is \ (50 \ % \), SO \ (f (I, A), F (I, B )\) the difference between them is reduced by \ (50 \ % \), which is equivalent to \ (f (I, a) = f (I, B) = (f (I,) + f (I, B) \ Div 2 \). Similarly, \ (f (a, I) = f (B, I) = (f (a, I) + f (B, I) \ Div 2 \). The final backward direction is expected, that is, \ (\ Sigma [I <j] f (I, j) \ times 1 \), that is \ (\ Sigma [I <j] f (I, j )\).

Let's say a few more things.In fact, as long as we come up with \ (f (I, j) \), everything is simple. But how many people can think of this method? There is no similar situation as a reference. mastering this question can provide experience for similar questions (after all, there is no similar question ). The next time I saw this question with a large amount of thinking, I still cannot figure it out. The active thinking in \ (OI \) still plays a major role!

AC code

#include<bits/stdc++.h>#define RG registerusing namespace std;int n,m,a[1005];double ans,f[1005][1005];int read(){    RG int re=0;RG char ch=getchar();    while(!isdigit(ch)) ch=getchar();    while(isdigit(ch)) re=(re<<3)+(re<<1)+ch-'0',ch=getchar();    return re;}int main(){    n=read(),m=read();    for(RG int i=1;i<=n;i++) a[i]=read();    for(RG int i=1;i<=n;i++)        for(RG int j=i+1;j<=n;j++)            if(a[i]>a[j]) f[i][j]=1.0;            else f[j][i]=1.0;    while(m--)    {        RG int x=read(),y=read();        if(x==y) continue;        for(RG 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;            f[x][i]=f[y][i]=(f[x][i]+f[y][i])/2;        }        f[x][y]=f[y][x]=0.5;    }    for(RG int i=1;i<=n;i++)        for(RG int j=i+1;j<=n;j++)            ans+=f[i][j];    printf("%.8f",ans);    return 0;}

Codeforces 258d little elephant and broken sorting (expected)