hdu3853 LOOPS 概率dp求期望

LOOPS

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)

Total Submission(s): 1790    Accepted Submission(s): 698

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

2011 Invitational Contest Host by BUPT

 

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题意：给定一个n*m的方格，给出相应( i , j ) 原地、往下 、往右走的概率，每走一格花费魔法值2，求从起始点（1,1）到（n,m）花费魔法值的期望

思路：设dp[ i ][ j ]为从（i , j）出发到达目标状态（n , m）所需魔法值的期望。dp[ i ][ j ]有3种转移方式：
（1）dp[ i ][ j ] -------> dp[ i ][ j ],转移概率为p1;
（2）dp[ i ][ j ] -------> dp[ i ][ j + 1 ] ,转移概率为p2;
（3）dp[ i ][ j ] -------> dp[ i + 1 ][ j ] ,转移概率为p3;
又因为总期望为子期望的加权和，加权因子为子期望的转移概率，所以得到：
dp[ i ][ j ]= p1 * ( dp[ i ][ j ] + 2 ) + p2 * ( dp[ i ][ j + 1 ] + 2 ) + p3 * ( dp[ i + 1 ][ j ] + 2) ;
化简得：
dp[ i ][ j ]=( 2 + p2 * dp[ i ][ j + 1 ] + p3 * dp[ i + 1 ][ j ] ) / ( 1 - p1) ;

注意：可能存在p1=1.0的时候，即说明该点不被访问（因为题目说期望有范围，而此时该点期望为inf）,所以dp[ i ][ j ]要置0。详见代码：

#include
#include
#include
#include
using namespace std;
const double eps=1e-5;
const int MAXN=1000+100;
double p[MAXN][MAXN][3],dp[MAXN][MAXN];

int n,m;
int main()
{
    //freopen("text.txt","r",stdin);
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                for(int k=0;k<3;k++)
                    scanf("%lf",&p[i][j][k]);
        memset(dp,0,sizeof(dp));
        for(int i=n;i>=1;i--)
            for(int j=m;j>=1;j--)
            {
                if(i==n && j==m)
                    continue;
                if(fabs(1-p[i][j][0])