HDOJ 3583 LOOPS（期望DP）

LOOPS

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 20.00 0.50 0.50    0.50 0.00 0.500.50 0.50 0.00    1.00 0.00 0.00

 

Sample Output

6.000

题目大意：一个萌萌哒小女孩被困在一个r行c列的迷宫里啦，她现在位于点(1,1)处，迷宫的出口位于点(r,c)。在每一个点(i,j)都有一定的概率移动到(i + 1,j)、(i,j + 1)或者原地不动。每一次移动萌萌哒小女孩都要花费2单位魔力，求这个萌萌哒小女孩逃出迷宫所消耗魔力的期望值。
解题思路：又是求期望，时刻提醒自己正向推概率，反向推期望。首先用一个结构体保存每个格子移动方向的概率，定义dp[i][j]为处于格子(i,j)时移动到格子(r,c)时所需要的魔力。显然，dp[r][c] = 0,dp[1][1]为我们所求。状态转移方程也比较简单

dp[i][j] =  dp[i][j] * maze[i][j].r_c + dp[i][j + 1] * maze[i][j].r_c1 + dp[i + 1][j] * maze[i][j].r1_c + 2

化简后得到如下方程

dp[i][j] =  (dp[i][j + 1] * maze[i][j].r_c1 + dp[i + 1][j] * maze[i][j].r1_c + 2) / (1 - maze[i][j].r_c)

根据此方程，反向推导dp[i][j]即可求得结果。

代码如下：

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;const double EPS = 1e-6;const int maxn = 1005;struct grid{    double r_c;    double r_c1;    double r1_c;};grid maze[maxn][maxn];double dp[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",&maze[i][j].r_c,&maze[i][j].r_c1,&maze[i][j].r1_c);            }        }        double ans;        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 - maze[i][j].r_c) < EPS){                    dp[i][j] = 0;                    continue;                }                dp[i][j] =  (dp[i][j + 1] * maze[i][j].r_c1 + dp[i + 1][j] * maze[i][j].r1_c + 2) / (1 - maze[i][j].r_c);            }        }        printf("%.3f\n",dp[1][1]);    }    return 0;}