Hdu 3853 LOOPS（概率dp）

题目链接

LOOPS

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2726    Accepted Submission(s): 1126

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
  

 

题意：有一个r*c的格子，初始的时候在左上角，要找到右下角。已知耗费2点魔法，从每个格子走到它右边的格子的概率，走到它下边的格子的概率，在该格子不动的概率。求走到右下角要花费的魔法的期望值？

题解：用dp[i][j] 表示当前在格子（i，j），从该格子走到目标格子要花费的魔法的期望值。

dp[r][c]=0;

设格子（i，j）原地不动的概率为p1，往右走的概率为p2，往下走的概率为p3.转移就是

dp[i][j]=2+p1*dp[i][j]+p2*dp[i][j+1]+p3*dp[i+1][j];

化简得：

dp[i][j]=(2+p2*dp[i][j+1]+p3*dp[i+1][j])/(1-p1)。

注意：题目说明答案保证不超过1000000。但是有些格子可能本来就不能到达的，所以p1可能为1。当p1等于1的时候，计算机计算的值是有问题的，所以我们要特判p1为1的情况。

代码如下：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<string>
#include<stack>
#include<math.h>
#define nn 1100
#define inff 0x3fffffff
#define eps 1e-8
#define mod 1000000007
typedef long long LL;
const LL inf64=LL(inff)*inff;
using namespace std;
int n,m;
double dp[nn][nn];
double p1[nn][nn],p2[nn][nn],p3[nn][nn];
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf("%lf%lf%lf",&p1[i][j],&p2[i][j],&p3[i][j]);
            }
        }
        memset(dp,0.0,sizeof(dp));
        for(i=n;i>=1;i--)
        {
            for(j=m;j>=1;j--)
            {
                if(i==n&&j==m)
                    continue;
                if(1.0-p1[i][j]>eps)
                    dp[i][j]=(2.0+dp[i+1][j]*p3[i][j]+dp[i][j+1]*p2[i][j])/(1.0-p1[i][j]);
            }
        }
        printf("%.3lf\n",dp[1][1]);
    }
    return 0;
}