[dp] hdu3853 loops

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本文深入探讨了Puella Magi系列中，魔法少女Homura如何运用智慧与魔法，试图通过复杂的迷宫逃脱，面对Boss Incubator的诡计。通过概率动态规划，计算出她逃脱所需的期望魔法能量。

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 c_th 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

题目分析

    此题为概率dp，求从起点到终点所花费的期望。
    设f[i][j]为从(i,j)点走到终点所花费的期望，a[i][j],b[i][j],c[i][j]分别是留在原地、向右走、向下走的概率，则可以得到：
    f[i][j]=a[i][j]*f[i][j]+b[i][j]*f[i][j+1]+c[i][j]*f[i+1][j]+2
    即为Σ往某个方向走的概率*走过去之后的期望花费最后再算上本次花费的magic power 2。
    移项可得：
    f[i][j]=(a[i][j]*f[i][j]+b[i][j]*f[i][j+1]+c[i][j]*f[i+1][j]+2)/(1-a[i][j])
    另由状态转移方程可知，循环变量要从大到小

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
const int maxn=1001;
double f[maxn][maxn];
double a[maxn][maxn],b[maxn][maxn],c[maxn][maxn];
const double tor=0.0000001;
int main()
{
    int m,n;
    while(cin>>m>>n)
    {
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                scanf("%lf%lf%lf",&a[i][j],&b[i][j],&c[i][j]);
        memset(f,0,sizeof f);
        for(int i=m-1;i>=0;i--)
            for(int j=n-1;j>=0;j--)
            {
                if(((a[i][j]-1)<tor)&&((1-a[i][j])<tor))
                    continue;
                f[i][j]=(b[i][j]*f[i][j+1]+c[i][j]*f[i+1][j]+2)/(1-a[i][j]);
            }
        printf("%.3lf\n",f[0][0]);
    }

    return 0;
}