hdu 3853 LOOPS

Title:

Given an n * m lattice, a person starts at (1,1) and has to go to (n,m).

Suppose at (x,y), every time she can spend 2 magic energy, the probability of p1 goes to (x,y), the probability of p2 to go to (x,y+1), the probability of p3 to go to (x +1,y), guaranteeing p1+p2+p3 = 1.

Ask what is the expected cost of walking from (1,1) to (n,m).

Ideas:

Short answer probability dp, inverse.

dp[i][j] = (dp[i][j+1] * p2 + dp[i+1][j] * p3) / (1 - p1)。

But pay attention to skip when p1 == 1 .

Code:

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 const int N = 1005;
 6 double dp[N][N];
 7 double a[N][N][3];
 8 int main()
 9 {
10     int n,m;
11     while (scanf("%d%d",&n,&m) != EOF)
12     {
13         memset(dp,0,sizeof(dp));
14         for (int i = 1;i <= n;i++)
15         {
16             for (int j = 1;j <= m;j++)
17             {
18                 for (int k = 0;k < 3;k++)
19                 {
20                     scanf("%lf",&a[i][j][k]);
21                 }
22             }
23         }
24         dp[n][m] = 0;
25         for (int i = n;i >= 1;i--)
26         {
27             for (int j = m;j >= 1;j--)
28             {
29                 if (i == n && j == m) continue;
30                 if (fabs(1.0-a[i][j][0]) < 1e-8) continue;
31                 dp[i][j] = (a[i][j][1] * dp[i][j+1] + a[i][j][2] * dp[i+1][j] + 2.0) / (1.0-a[i][j][0]);
32             }
33         }
34         printf("%.3f\n",dp[1][1]);
35     }
36     return 0;
37 }