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.
InputThe first line contains two integers R and C (2 <= R, C <= 1000).
The following R lines, each contains C*3 real numbers, at 2 decimal places. Every three numbers make a group. The first, second and third number of the cth group of line r represent the probability of transportation to grid (r, c), grid (r, c+1), grid (r+1, c) of the portal in grid (r, c) respectively. Two groups of numbers are separated by 4 spaces.
It is ensured that the sum of three numbers in each group is 1, and the second numbers of the rightmost groups are 0 (as there are no grids on the right of them) while the third numbers of the downmost groups are 0 (as there are no grids below them).
You may ignore the last three numbers of the input data. They are printed just for looking neat.
The answer is ensured no greater than 1000000.
Terminal at EOF
OutputA real number at 3 decimal places (round to), representing the expect magic power Homura need to escape from the LOOPS.
Sample Input
2 2 0.00 0.50 0.50 0.50 0.00 0.50 0.50 0.50 0.00 1.00 0.00 0.00
Sample Output
6.000
题意: R 行 C 列的迷宫，起点 (1,1) ,终点(R,C) ,然后给出每个格子的留在原地的概率，向右的概率，向下的概率，走一步 2 的魔法，问能逃出去的耗费的魔法期望
题解:一般做法，dp [i][j] 代表从(i,j)逃出去的魔法消耗期望，值得注意的是如果这点停在原地概率为 1 的话，这点的 dp[][]值应该为 0 ，因为不能进去，进去就是死，要求的是能逃得出的魔法消耗期望
1 #include <iostream>
2 #include <stdio.h>
3 #include <string.h>
4 #include <math.h>
5 using namespace std;
6 #define MAXN 1005
7 const double eps = 1e-8;
8
9 struct Pro
10 {
11 double s;
12 double r;
13 double d;
14 }p[MAXN][MAXN];
15
16 int n,m;
17 double dp[MAXN][MAXN];
18
19 int main()
20 {
21 while (scanf("%d%d",&n,&m)!=EOF)
22 {
23 for (int i=1;i<=n;i++)
24 for (int j=1;j<=m;j++)
25 scanf("%lf%lf%lf",&p[i][j].s,&p[i][j].r,&p[i][j].d);
26 memset(dp,0,sizeof(dp));
27 for (int i=n;i>=1;i--)
28 for (int j=m;j>=1;j--)
29 {
30 if (i==n&&m==j) continue;
31 if (1-p[i][j].s<eps) continue; //如果在这点必死，则dp值继续保持 0 代表不能来
32 dp[i][j]=(p[i][j].r*dp[i][j+1]
33 +p[i][j].d*dp[i+1][j]
34 +2)/(1-p[i][j].s);
35 }
36 printf("%.3lf\n",dp[1][1]);
37 }
38 return 0;
39 }