题意

D. Broken robottime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

You received as a gift a very clever robot walking on a rectangular board. Unfortunately, you understood that it is broken and behaves rather strangely (randomly). The board consists of N rows and M columns of cells. The robot is initially at some cell on the i-th row and the j-th column. Then at every step the robot could go to some another cell. The aim is to go to the bottommost (N-th) row. The robot can stay at it's current cell, move to the left, move to the right, or move to the cell below the current. If the robot is in the leftmost column it cannot move to the left, and if it is in the rightmost column it cannot move to the right. At every step all possible moves are equally probable. Return the expected number of step to reach the bottommost row.

Input

On the first line you will be given two space separated integers N and M (1 ≤ N, M ≤ 1000). On the second line you will be given another two space separated integers i and j (1 ≤ i ≤ N, 1 ≤ j ≤ M) — the number of the initial row and the number of the initial column. Note that, (1, 1) is the upper left corner of the board and (N, M) is the bottom right corner.

Output

Output the expected number of steps on a line of itself with at least 4 digits after the decimal point.

ExamplesInputCopy

10 10
10 4

OutputCopy

0.0000000000

InputCopy

10 14
5 14

OutputCopy

18.0038068653

分析

参照M_sea的博客。

设\(f[i][j]\)表示\((i,j)\)到最后一排的移动步数的期望。

容易得出：

\(f[i][1]=\frac{1}{3}(f[i][1]+f[i][2]+f[i-1][1])+1\)

\(f[i][m]=\frac{1}{3}(f[i][m]+f[i][m-1]+f[i-1][m])+1\)

\(f[i][j]=\frac{1}{4}(f[i][j]+f[i][j-1]+f[i][j+1]+f[i-1][j]\quad (1<j<m)\)

发现列与列的转移是有后效性的。

所以用有后效性DP的基本方法——DP套高斯消元。

但是发现系数矩阵中每行只有几个数要消，而且非常有规律。所以可以\(O(m)\)解。

细节见代码。注意要特判\(m=1\)的情况。

代码

第一次打特殊的高斯消元……学到了呀。

#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;

co int N=1002;
int n,m,x,y;
double f[N][N],d[N][N];
void work(){
    for(int i=1;i<=m;++i){
        double w=1/d[i][i];
        d[i][i]*=w,
        d[i][m+1]*=w;
        if(i==m) break;
        d[i][i+1]*=w;
        w=d[i+1][i]/d[i][i];
        d[i+1][i]-=w*d[i][i],
        d[i+1][i+1]-=w*d[i][i+1],
        d[i+1][m+1]-=w*d[i][m+1];
    }
    for(int i=m-1;i;--i)
        d[i][m+1]-=d[i+1][m+1]*d[i][i+1];
}
int main(){
//  freopen(".in","r",stdin),freopen(".out","w",stdout);
    read(n),read(m),read(x),read(y);
    for(int i=n-1;i>=x;--i){
        d[1][1]=d[m][m]=-2/3.0,
        d[1][2]=d[m][m-1]=1/3.0;
        for(int j=2;j<m;++j)
            d[j][m+1]=-f[i+1][j]/4.0-1,
            d[j][j]=-3/4.0,
            d[j][j-1]=d[j][j+1]=1/4.0;
        if(m==1) d[1][1]=-1/2.0;
        d[1][m+1]=-f[i+1][1]/3.0-1;
        d[m][m+1]=-f[i+1][m]/3.0-1;
        if(m==1) d[m][m+1]=-f[i+1][m]/2.0-1;
        work();
        for(int j=1;j<=m;++j) f[i][j]=d[j][m+1];
    }
    printf("%lf\n",f[x][y]);
    return 0;
}