LOOPS
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 6423 Accepted Submission(s): 2572
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
Recommend
chenyongfu
题解：刚开始左概率dp，就从这个最简单的概率dp开始吧。
题目要求从左上角到右下角消耗的魔法的期望。如果要找一个位置的期望与它左边和上面的的期望的关系
会比较困难，所以，就反着来找，找一个位置的期望与它右边和下面的的期望的关系，设dp[i][j]为(i,j)到(r,c)
消耗魔法的期望：
有dp[i][j]=p1*dp[i][j]+p2*dp[i][j+1]+p3*dp[i+1][j]+2，化简可得：
dp[i][j]=（p2*dp[i][j+1]+p3*dp[i+1][j]+2）/(1-p1);
如果p1=1则无法到达终点,题目说答案不小于1000000
则除了最后一点，就没有p1=1的点；或者不能到p1=1的点
#include<cstdio> #include<cstring> #include<cmath> const double eps=1e-7; struct node { double a,b,c; } w[1005][1005]; double dp[1005][1005]; int main() { int r,c; while(~scanf("%d%d",&r,&c)) { for(int i=1; i<=r; i++) for(int j=1; j<=c; j++) scanf("%lf%lf%lf",&w[i][j].a,&w[i][j].b,&w[i][j].c); dp[r][c]=0; for(int i=r; i>=1; i--) for(int j=c; j>=1; j--) if(fabs(1-w[i][j].a)>eps) dp[i][j]=(w[i][j].b*dp[i][j+1]+w[i][j].c*dp[i+1][j]+2)/(1-w[i][j].a); printf("%.3lf ",dp[1][1]); } return 0; }