HDU 3853 LOOPS 概率dp

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我是靠谱客的博主天真金毛，最近开发中收集的这篇文章主要介绍HDU 3853 LOOPS 概率dp，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

E - LOOPS

Time Limit:5000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3853

Appoint description:

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

晓美焰要从（1，1）到（n，n)在每个点都有3种可能不走或向右或向下走每次花费2个魔法值

求花费的魔法值的期望求期望都说倒着推

那么每个点的下一个状态可能是右边或下面

那么 dp[i][j]=(dp[i][j+1]*右边的概率+dp[i+1][j]*下面的概率+花费)/（右边+下面的概率）

ACcode：

#include <cstdio>
#include <cstring>
#include <cmath>
#define maxn 1002
using namespace std;
double dp[maxn][maxn],a[maxn][maxn],b[maxn][maxn],c[maxn][maxn];
int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;++i)for(int j=1;j<=m;++j)scanf("%lf%lf%lf",&a[i][j],&b[i][j],&c[i][j]);
for(int i=n;i>=1;--i)for(int j=m;j>=1;--j)
if((i==n&&j==m)||(fabs(1-a[i][j])<1e-7))continue;
else dp[i][j]=(dp[i][j+1]*b[i][j]+dp[i+1][j]*c[i][j]+2)/(b[i][j]+c[i][j]);
printf("%.3lfn",dp[1][1]);
}
}
/*
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
*/