Hdu--4487--dp

DP is the most interesting. Because it never has a fixed routine.

The most boring is the template problem.

I got a hole in the problem.   Test instructions read the wrong. Here's a more interesting question was what's the expected rightmost position you'll attain during the walk.

That is, the rightmost position that we can reach in each walk, at first, as far as the right-hand position can be reached after n-step = =

This is a very obvious linear expectation DP.

A total of 2 state transitions for dp[x,y,z] go to step x the current position is Y at this point in the process of X-step, the rightmost distance is Z

Then Y and Z have a total of 3 size relationships.

Y<z

Dp[x&1][y][z] = dp[(x-1) &1][y+1][z] * L + dp[(x-1) &1][y][z] * M + dp[(x-1) &1][y-1][z] * R;

First you have to be clear dp[x,y,z] I was transferred by this state of dp[x-1,u,v, so it's guaranteed that even if I walk a step forward at U, it's not going to be more than V.

Y==z

Dp[x&1][y][z] = dp[(x-1) &1][y-1][z-1] * r + dp[(x-1) &1][y][z] * M + dp[(x-1) &1][y-1][z] * R;

This is probably the first time I have reached the z position, or I stopped at the z position in the last step, or I got to the top of Z, but the last step to the left, this step to the right, or the furthest distance is Z.

Y<z

Because I'm now in the Y position, so far from the right, Z must be at least equal to Y. < left to take the origin processing >

As to whether this scrolling array is optimized or not, it doesn't matter. Data is small 100

1#include <iostream>2#include <cstring>3#include <algorithm>4 using namespacestd;5 6 intN;7 DoubleL, R, M;8 Const intSize = the;9 Doubledp[2][size*2][size*2];//Dp[i,j,k] Go to step I at the time of the J this position to reach the far right distance is KTen  One voidInit () A { -Memset (DP,0,sizeof(DP)); -dp[0][size][size] =1; the } -  - voidSolve () - { +      for(inti =1; I<=n; i++ ) -     { +          for(intj = size-i; I<=size+i; J + + ) A         { at              for(intk = max (j,size); K<=size+i; k++ ) -             { -                 if(j==k)//The first time I walked to J, I walked to K and walked back. -                 { -dp[i&1][j][k] = dp[(i-1) &1][j-1][k-1] * R + dp[(i-1) &1][J][K] * M + dp[(i-1) &1][j-1][K] *R; -                 } in                 Else//case of k > J -                 { todp[i&1][j][k] = dp[(i-1) &1][j+1][K] * L + dp[(i-1) &1][J][K] * M + dp[(i-1) &1][j-1][K] *R; +                 } -             } the         } *     } $ }Panax Notoginseng  - intMain () the { +     intT, M; A     Doubleans; thescanf"%d",&t); +      while(t-- ) -     { $ init (); $scanf"%d%d%lf%lf",&m,&n,&l,&R); -M =1LR; - solve ();  theAns =0; -          for(intj = size-n; J<=size+n; J + + )Wuyi         { the              for(intk = size; K<=size+n; k++ ) -             { WuAns + = dp[n&1][j][k] * (K-size); -             } About         } $printf"%.4lf\n", ans); -     } -     return 0; -}

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Hdu--4487--dp