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:

1. Up (from (x, y) to (x, y + 1)) with probability U.
2. Right (from (x, y) to (x + 1, y)) with probability R.
3. Down (from (x, y) to (x, y - 1)) with probability D.
4. 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[410][410][210],dp2[410][410][210],dp3[410][410][210],dp4[410][410][210];
int vis1[410][410][210],vis2[410][410][210],vis3[410][410][210],vis4[410][410][210];
double maxx(int x,int R,int step)
{
    if(vis1[x][R][step]) return dp1[x][R][step];
    if(step==0) return R;
    double res=0;
    res+=maxx(x,R,step-1)*(u+d);
    res+=maxx(x+1,max(R,x+1),step-1)*r;
    res+=maxx(x-1,R,step-1)*l;
    vis1[x][R][step]=1; 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==0) return L;
    double res=0;
    res+=minx(x,L,step-1)*(u+d);
    res+=minx(x+1,L,step-1)*r;
    res+=minx(x-1,min(L,x-1),step-1)*l;
    vis2[x][L][step]=1; 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==0) return U;
    double res=0;
    res+=maxy(y,U,step-1)*(l+r);
    res+=maxy(y+1,max(U,y+1),step-1)*u;
    res+=maxy(y-1,U,step-1)*d;
    vis3[y][U][step]=1; 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==0) return D;
    double res=0;
    res+=miny(y,D,step-1)*(l+r);
    res+=miny(y+1,D,step-1)*u;
    res+=miny(y-1,min(D,y-1),step-1)*d;
    vis4[y][D][step]=1; dp4[y][D][step]=res; 
    return res;
}
int main()
{
    int N; 
    cin>>N>>u>>r>>d>>l;
    ans+=maxx(201,201,N);
    ans-=minx(201,201,N);
    ans+=maxy(201,201,N);
    ans-=miny(201,201,N);
    printf("%.7lf\n",ans);
    return 0;
}