LOOPS
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 649 Accepted Submission(s): 267
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
比較簡單的概率DP了。很容易寫出方程去遞推出答案。
注意特殊情況要判斷下。
/* HDU 3853 解析： 設dp[i][j]表示(i,j)到(R,C)需要消耗的能量 則： dp[i][j]=p1[i][j]*dp[i][j]+p2[i][j]*dp[i][j+1]+p3[i][j]*dp[i+1][j]+2; 化簡得到： dp[i][j]=p2[i][j]*dp[i][j+1]/(1-p1[i][j])+p3[i][j]*dp[i+1][j]/(1-p1[i][j])+2/(1-p1[i][j]); 注意一種情況就是p1[i][j]==1的情況。 題目只是保證答案小於1000000.但是有的點可能永遠都不可能到達的。 所以這樣的點出現p1[i][j]是允許的。 否則就會WA了。 */ #include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #include<math.h> using namespace std; const int MAXN=1010; const double eps=1e-5; double dp[MAXN][MAXN]; double p1[MAXN][MAXN]; double p2[MAXN][MAXN]; double p3[MAXN][MAXN]; int main() { int R,C; while(scanf("%d%d",&R,&C)!=EOF) { for(int i=1;i<=R;i++) for(int j=1;j<=C;j++) scanf("%lf%lf%lf",&p1[i][j],&p2[i][j],&p3[i][j]); dp[R][C]=0; for(int i=R;i>=1;i--) for(int j=C;j>=1;j--) { if(i==R&&j==C)continue; if(fabs(1-p1[i][j])<eps)continue; dp[i][j]=p2[i][j]/(1-p1[i][j])*dp[i][j+1]+p3[i][j]/(1-p1[i][j])*dp[i+1][j]+2/(1-p1[i][j]); } printf("%.3lf\n",dp[1][1]); } return 0; }