SPOJ：Robot（数学期望）

There is a robot on the 2D plane. Robot initially standing on the position (0, 0). Robot can make a 4 different moves:

Up (from (x, y) to (x, y + 1)) with probability U.
Right (from (x, y) to (x + 1, y)) with probability R.
Down (from (x, y) to (x, y - 1)) with probability D.
Left (from (x, y) to (x - 1, y)) with probability L.

After moving N times Robot gets points.

Let x1 be the smallest coordinate in X-axis, that Robot reached in some moment.
Let x2 be the largest coordinate in X-axis, that Robot reached in some moment.
Let y1 be the smallest coordinate in Y-axis, that Robot reached in some moment.
Let y2 be the largest coordinate in Y-axis, that Robot reached in some moment.

Points achieved by Robot equals to x2 - x1 + y2 - y1.

Given N, U, R, D, L. Calculate expected value of points that Robot achieved after N moves.

Input

First line: One interger N (1 ≤ N ≤ 200).

Second line: Four real numbers U, R, D, L (U + R + D + L = 1, 0 ≤ U, R, D, L ≤ 1) with maximum of 6 numbers after dot.

Output

One number: expected value of points achieved by Robot. The answer will be considered correct if its relative or absolute error does not exceed 10-6.

Example 1

Input:

2

0.100000 0.200000 0.300000 0.400000

Output:

1.780000

Example 2

Input:

3

0.25 0.25 0.25 0.25

Output:

2.375000

题意：二维平面上一个机器人，给出机器人走的步数，已知走上下左右的概率；机器人走到最右边是Xmax，最左边是Xmin，上下同理。问机器人Xmax-Xmin+Ymax-Ymin的期望。

思路：万万没想到，4个方向是分开求，开始一直在想怎么整体求。。。。分开求的话就不难想了，分别求出X方向的最大最小，Y方向的最大最小，搜索一下。。。自己看代码。。。但是注意搜索会超时，注意记忆化。。。

（总之，是个不错的题！

#include<bits/stdc++.h>

using namespace std;

double u,d,l,r,ans;

double dp1[][][],dp2[][][],dp3[][][],dp4[][][];

int vis1[][][],vis2[][][],vis3[][][],vis4[][][];

double maxx(int x,int R,int step)

{

    if(vis1[x][R][step]) return dp1[x][R][step];

    if(step==) return R;

    double res=;

    res+=maxx(x,R,step-)*(u+d);

    res+=maxx(x+,max(R,x+),step-)*r;

    res+=maxx(x-,R,step-)*l;

    vis1[x][R][step]=; dp1[x][R][step]=res;

    return res;

}

double minx(int x,int L,int step)

{

    if(vis2[x][L][step]) return dp2[x][L][step];

    if(step==) return L;

    double res=;

    res+=minx(x,L,step-)*(u+d);

    res+=minx(x+,L,step-)*r;

    res+=minx(x-,min(L,x-),step-)*l;

    vis2[x][L][step]=; dp2[x][L][step]=res;

    return res;

}

double maxy(int y,int U,int step)

{

    if(vis3[y][U][step]) return dp3[y][U][step];

    if(step==) return U;

    double res=;

    res+=maxy(y,U,step-)*(l+r);

    res+=maxy(y+,max(U,y+),step-)*u;

    res+=maxy(y-,U,step-)*d;

    vis3[y][U][step]=; dp3[y][U][step]=res;

    return res;

}

double miny(int y,int D,int step)

{

    if(vis4[y][D][step]) return dp4[y][D][step];

    if(step==) return D;

    double res=;

    res+=miny(y,D,step-)*(l+r);

    res+=miny(y+,D,step-)*u;

    res+=miny(y-,min(D,y-),step-)*d;

    vis4[y][D][step]=; dp4[y][D][step]=res;

    return res;

}

int main()

{

    int N;

    cin>>N>>u>>r>>d>>l;

    ans+=maxx(,,N);

    ans-=minx(,,N);

    ans+=maxy(,,N);

    ans-=miny(,,N);

    printf("%.7lf\n",ans);

    return ;

}