LOOPS
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 6457 Accepted Submission(s): 2592
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.
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
Source
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/*状态可以定义为dp[i][j]，代表从（i，j）到（r，c）所花费魔法值的期望。然后我们需要考虑这样的状态之间能否正确的转化，利用数学期望的定义以及其线性性不难写出如下转移方程：dp[i][j] = p[i][j][1]*dp[i][j] + p[i][j][2]*dp[i][j+1] + p[i][j][3]*dp[i+1][j] + 2（其中p[i][j][k]代表在点（i，j）选择第k种走法的概率）再化简一下：dp[i][j] = （p[i][j][2]*dp[i][j+1] + p[i][j][3]*dp[i+1][j] + 2）/（1-p[i][j][1]）。最后，需要确定边界，很明显，dp[r][c]=0，因为当在点（r，c）时，他不需要花费魔法值就可以到达（r，c）dp[1][1]为答案 */#include#include #include #define N 1005using namespace std;double dp[N][N],p[N][N][3];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++) for(int k=1;k<=3;k++) scanf("%lf",&p[i][j][k]); dp[r][c]=0; for(int i=r;i>0;i--) { for(int j=c;j>0;j--) { if(p[i][j][1]==1|| (i==r)&&j==c) continue; dp[i][j]=(p[i][j][2]*dp[i][j+1]+p[i][j][3]*dp[i+1][j]+2)/(1-p[i][j][1]); } } printf("%.3lf\n",dp[1][1]); } return 0;}