hud 3853 LOOPS （dp求概率期望）

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

大致题意：一个马猴烧酒（魔法少女）被困在一个魔法矩阵中，矩阵的规格是r*c，起始点是（1，1），出口是（r，c）,每个格子中有一个魔法阵，需要花费2个法力点发动，马猴烧酒通过魔法阵可能从当前格子（i，j）移动到下面（i+1，j）格子或者左边（i，j+1）格子，或者停留在原地。问达到出口（r，c）所需花费的期望法力值。 首先输入r，c，接下来输入每个格子停留在原地，到右边格子和到下面格子的三种情况的概率。

思路：假设dp[i][j] 表示从（i，j）点到终点（r，c）还需花费的法力值的期望
mp[i][j].(now,down,right)表示点（i，j）的三种情况的概率

那么 dp[r][c]=0
由： dp[i][j]=dp[i][j]*mp[i][j].now+dp[i+1][j]*mp[i][j].down+dp[i][j+1]*mp[i][j].right+2

推得： dp[i][j]=(dp[i][j+1]*mp[i][j].right+dp[i+1][j]*mp[i][j].down+2)/(1-mp[i][j].now);

dp[1][1]即为答案

注意当mp[i][j].now==1时，马猴烧酒会一直停留在当前的格子

代码如下

#include<iostream>#include<cstring>#include<cmath>using namespace std;struct node{    double now,right,down;//存三种情况的概率};node mp[1010][1010];double dp[1010][1010];  int main(){    int r,c;    while(scanf("%d%d",&r,&c)!=EOF)    {    memset(dp,0,sizeof(dp));    for(int i=1;i<=r;i++)    for(int j=1;j<=c;j++)    {        scanf("%lf%lf%lf",&mp[i][j].now,&mp[i][j].right,&mp[i][j].down);    }    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(mp[i][j].now==1.0)        continue;        dp[i][j]=(dp[i][j+1]*mp[i][j].right+dp[i+1][j]*mp[i][j].down+2)/(1-mp[i][j].now);    }    printf("%.3lf\n",dp[1][1]);    }    return 0; }