E[n][m] = 0;
三种运动的概率分别是a,b,c
E[i][j] = a*E[i][j] + b*E[i][j+1] + c*E[i+1][j]
->E[i][j] = ( b*E[i][j+1] + c*E[i+1][j] )/(1-a)
如果1-a近似到0，那么说明只能到它本身，所以它不能到达终点，所以它的期望是无穷，也就是不能到达，要进行特判，赋值为０
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define MAX 1007
#define eps 1e-8
using namespace std;
double dp[MAX][MAX];
double p[MAX][MAX][4];
int n,m;
void dfs ( int x , int y )
{
if ( x== n && y == m )
{
dp[x][y] = 0.0;
return;
}
if ( dp[x][y] > -0.5 ) return;
dp[x][y] = 2.0;
if ( y < m && dp[x][y+1] < -0.5 ) dfs ( x , y+1 );
dp[x][y] += dp[x][y+1]*p[x][y][2];
if ( x < n && dp[x+1][y] < -0.5 ) dfs ( x+1 , y );
dp[x][y] += dp[x+1][y]*p[x][y][3];
if ( fabs ( 1.0 - p[x][y][1] ) < eps ) dp[x][y] = 0.0;
else dp[x][y] /= ( 1.0 - p[x][y][1] );
}
int main ( )
{
while ( ~scanf ( "%d%d" , &n , &m ) )
{
for ( int i = 1 ; i <= n ; i++ )
for ( int j = 1 ; j <= m ; j++ )
for ( int t = 1 ; t < 4 ; t++ )
scanf ( "%lf" , &p[i][j][t] );
for ( int i = 0 ; i <= n ; i++ )
for ( int j = 0 ; j <= m ; j++ )
dp[i][j] = -1.0;
dfs ( 1 , 1 );
printf ( "%.3lf\n" , dp[1][1] );
}
}
