Description
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 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
Sample Input
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
题意：
有一个迷宫r行c列,开始点在[1,1]现在要走到[r,c]
对于在点[x,y]可以打开一扇门走到[x,y]或者[x+1,y]或者[x,y+1]
消耗2点魔力 问平均消耗多少魔力能走到[r,c]
分析：
输入r和c 随后r行c列 输入三个概率
假设dp[i][j]表示在点[i,j]到达[r,c]所需要消耗的平均魔力(期望)
则从dp[i][j]可以到达:
dp[i][j],dp[i+1,j],dp[i][j+1];
对应概率分别为： p1[i][j],p2[i][j],p3 [i][j]
由E(aA+bB+cC...)=aEA+bEB+cEC+...//包含状态A,B,C的期望可以分解子期望求解
得到dp[i][j]=p1[i][j]*dp[i][j]+p2[i][j]*dp[i+1][j]+p3[i][j]*dp[i][j+1]+2;
得出最终公式:dp[i][j]]=(p2[i][j]*dp[i+1][j]+p3[i][j]*dp[i][j+1]+2)/(1-p1[i][j])
注意分母为0的时候要特判一下
dp[i][j]表示从(i,j)走到(n,s)所需要消耗的魔力的期望值。
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
double f[][];
double p1[][],p2[][],p3[][];
int main()
{
int n,s;
while(scanf("%d%d",&n,&s)!=EOF)
{
for(int i=;i<=n;i++)
for(int j=;j<=s;j++)
scanf("%lf%lf%lf",&p1[i][j],&p2[i][j],&p3[i][j]);
memset(f,,sizeof(f));
for(int i=n;i>=;i--)
{
for(int j=s;j>=;j--)
{
if(i==n&&j==s)
continue;
if(p1[i][j]==1.00) //分母为0
continue;
f[i][j]=p2[i][j]*f[i][j+]+p3[i][j]*f[i+][j]+2.0;
f[i][j]/=(-p1[i][j]);
}
}
printf("%.3f\n",f[][]); //是f[1][1]，不是f[0][0]。
}
return ;
}