HDU 3853 概率dp

LOOPS

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 2337    Accepted Submission(s): 951

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

 

Recommend

chenyongfu

 

 

下面转自 http://blog.csdn.net/zhjchengfeng5/article/details/8611513

 

题目大意

    有一个人被困在一个 R*C(2<=R,C<=1000) 的迷宫中，起初他在 (1,1) 这个点，迷宫的出口是 (R,C)。在迷宫的每一个格子中，他能花费 2 个魔法值开启传送通道。假设他在 (x,y) 这个格子中，开启传送通道之后，有 p_lift[i][j] 的概率被送到 (x,y+1)，有 p_down[i][j] 的概率被送到 (x+1,y)，有 p_loop[i][j]  的概率被送到 (x,y)。问他到出口需要花费的魔法值的期望是多少。

做法分析

    令：f[i][j] 表示从 (i,j) 这个点到出口 (R,C) 花费的魔法值的期望。

    那么，我们有：

        f[i][j] = 2+p_loop[i][j]*f[i][j] + p_left[i][j]*f[i][j+1] + p_down[i][j]*f[i+1][j]

    移项可得：         (1-p_loop[i][j])*f[i][j] = 2+p_left[i][j](f[i][j+1] + p_down[i][j]*f[i+1][j]

    于是我们可以倒着递推了

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>

#define N 1005
#define M 15
#define mod 1000000007
#define mod2 100000000
#define ll long long
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

double a[N][N][3];
double dp[N][N];

int main()
{
    int i,j;
    int r,c;
    double p;
   // freopen("data.in","r",stdin);
    //scanf("%d",&T);
    //for(int cnt=1;cnt<=T;cnt++)
    //while(T--)
    while(scanf("%d%d",&r,&c)!=EOF)
    {
        for(i=1;i<=r;i++){
            for(j=1;j<=c;j++){
                scanf("%lf%lf%lf",&a[i][j][0],&a[i][j][1],&a[i][j][2]);
               // printf(" %.3lf %.3lf %.3lf
",a[i][j][0],a[i][j][1],a[i][j][2]);
            }
        }

        for(i=r;i>=1;i--){
            for(j=c;j>=1;j--){
                dp[i][j]=(2+a[i][j][1]*dp[i][j+1]+a[i][j][2]*dp[i+1][j]);
                if( (1-a[i][j][0])<1e-6){
                    dp[i][j]=0;
                }
                else
                    dp[i][j]/=(1-a[i][j][0]);
            }
        }
        printf("%.3f
",dp[1][1]);

    }

    return 0;
}