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.
Input
The 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
Output
A 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
題意：就是說一個人在一個R*C的格子上，他一開始在格子的（1,1）上，他最後要走到格子的（R,C），在後面的數據中，會給出他在每一個格子上留在原地、向下、向右走的概率，每走一步需要花費兩點代價。要走X步能到達（R,C），要求X的數學期望。
解析：我們不容易正着計算，但是我們可以逆推，我們可以看作從(R,C)點開始向回走，那麼對於每個點，可以從三種情況轉移過來：
- ①從(i,j)原地不動
- ②從(i,j+1)轉移而來
- ③從(i+1,j)轉移而來
我們用dp[i][j]表示倒着走到(i,j)所期望的步數，map[i][j][0~2]分別表示，(i,j)時停在原地、向右走、向下走的概率。
那麼根據這三種情況，我們就可以列出轉移方程：
dp[i][j]=dp[i][j]*map[i][j][0]+dp[i][j+1]*map[i][j][1]+dp[i+1][j]*map[i][j][2];
經過整理歸納，我們可以得出最後的式子：
dp[i][j]=(dp[i][j+1]*map[i][j][1]+dp[i+1][j]*map[i][j][2]+2)/(1-map[i][j][0]);
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int n,m;
double dp[1005][1005];
double map[1005][1005][4];
int make(double x){
double temp=x-1;
if(temp<0.00000){
temp=-temp;
}
if(temp<0.0000001){
return 1;
}
return 0;
}
int main(){
while(cin>>n>>m){
memset(dp,0,sizeof(dp));
memset(map,0,sizeof(map));
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
double x,y,z;
scanf("%lf%lf%lf",&x,&y,&z);
map[i][j][0]=x,map[i][j][1]=y,map[i][j][2]=z;
}
}
for(int i=n;i>=1;i--){
for(int j=m;j>=1;j--){
if(i==n&&j==m){
continue;
}
if(make(map[i][j][0])){
continue;
}
dp[i][j]=(dp[i][j+1]*map[i][j][1]+dp[i+1][j]*map[i][j][2]+2)/(1-map[i][j][0]);
}
}
printf("%.3lf\n",dp[1][1]);
}
return 0;
}