Problem Description
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.
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
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
Source
和POJ 2096的解法差不多（我的另外一篇博客），但是比那道题更容易理解，模型要更明显；
题目已经明确的说三个概率加起来一定等于1，所以很显然的得出概率的推导式，dp[i][j]可以转化成dp[i][j]，dp[i+1][j]以及dp[i][j+1]。
转移方程就能够写出：dp[i][j]=p1*dp[i][j]+p2*dp[i][j+1]+p3*dp[i+1][j]+2;
初始状态为dp[r][s]=0;
倒推求出dp[1][1]就是最后的答案；
注意：转移到自己的概率为1的情况要特殊注意，此时不能逃出迷宫！
#include<iostream> #include<cstdio> #include<vector> #include<set> #include<map> #include<string.h> #include<cmath> #include<algorithm> #include<queue> #include<stack> #define LL long long #define mod 1000000007 #define inf 0x3f3f3f3f using namespace std; double a[1010][1010][4]; double dp[1010][1010]; int main() { int r,s; while(~scanf("%d%d",&r,&s)) { memset(dp,0,sizeof(dp)); memset(a,0,sizeof(a)); for(int i=1;i<=r;i++) for(int j=1;j<=s;j++) { double p1,p2,p3; scanf("%lf%lf%lf",&p1,&p2,&p3); a[i][j][1]=p1; a[i][j][2]=p2; a[i][j][3]=p3; } for(int i=r;i>=1;i--) for(int j=s;j>=1;j--) { if(i==r&&j==s) continue; if(a[i][j][1]==1.0) continue; dp[i][j]=(a[i][j][2]*dp[i][j+1]+a[i][j][3]*dp[i+1][j]+2)/(1-a[i][j][1]); } printf("%.3lf ",dp[1][1]); } return 0; }