Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl).

Homura wants to help her friend Madoka save the world. But because of the plot of the Boss Incubator, she is trapped in a labyrinth called LOOPS.

The planform of the LOOPS is a rectangle of R*C grids. There is a portal in each grid except the exit grid. It costs Homura 2 magic power to use a portal once. The portal in a grid G(r, c) will send Homura to the grid below G (grid(r+1, c)), the grid on the right of G (grid(r, c+1)), or even G itself at respective probability (How evil the Boss Incubator is)!
At the beginning Homura is in the top left corner of the LOOPS ((1, 1)), and the exit of the labyrinth is in the bottom right corner ((R, C)). Given the probability of transmissions of each portal, your task is help poor Homura calculate the EXPECT magic power she need to escape from the LOOPS.

裸的期望DP

dp[i[j]表示当前在i,j位置，到达终点的期望

我们不难看出转移方程为

$dp[i][j]=dp[i][j]*p0+dp[i][j+1]*p1+dp[i+1][j]*p2$

解这个式子有两种方法

1.高斯消元

2.把右边的$dp[i][j]$除到左边

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<cmath>
 6 using namespace std;
 7 const int MAXN=2001;
 8 const int INF=0x7fffff;
 9 double dp[MAXN][MAXN];//走到i，j的期望 
10 struct node
11 {
12     double a,b,c;
13     void clear(){a=b=c=0.0;}
14 }map[MAXN][MAXN];
15 int n,m;
16 const double eps=1e-5;
17 int main()
18 {
19     dp[n][m]=0;
20     while(scanf("%d%d",&n,&m)!=EOF)
21     {
22         memset(dp,0.00,sizeof(dp));
23         dp[n][m]=0;
24         for(int i=1;i<=n;i++)
25             for(int j=1;j<=m;j++)
26                 map[i][j].clear();
27         for(int i=1;i<=n;i++)
28             for(int j=1;j<=m;j++)
29                 scanf("%lf%lf%lf",&map[i][j].a,&map[i][j].b,&map[i][j].c);
30         for(int i=n;i>=1;i--)
31             for(int j=m;j>=1;j--)
32             {
33                 if(i==n&&j==m)    continue;
34                 if(fabs(1.00-map[i][j].a)<eps)    continue;
35                 dp[i][j]=(map[i][j].b*dp[i][j+1]+map[i][j].c*dp[i+1][j]+2.00)/(1.0-map[i][j].a);
36             }    
37         printf("%.3lf\n",dp[1][1]);
38     }
39     
40     
41     return 0;
42 }