HDU 3853 LOOPS(概率dp)

畫船聽雨發表於2014-07-08
OOP

題目大意：迷宮是一個R*C的佈局，每個格子中給出停留在原地，往右走一個，往下走一格的概率，起點在(1,1)，終點在(R,C)，每走一格消耗兩點能量，求出最後所需要的能量期望

dp[i][j]表示：從(i,j)到終點的期望。所以還是dp[R][C] = 0;

每次走可以到達三種情況：1.自己的位置；2.i+1的位置；3.j+1的位置。

LOOPS

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1928    Accepted Submission(s): 765


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





 





#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 1000100
#define LL __int64
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 1010;

using namespace std;

double dp[maxn][maxn];
double p[maxn][maxn][3];

int main()
{
    int n, m;
    while(cin >>n>>m)
    {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                for(int k = 0; k < 3; k++)
                    scanf("%lf",&p[i][j][k]);
        memset(dp, 0, sizeof(dp));
        for(int i = n; i >= 1; i--)
        {
            for(int j = m; j >= 1; j--)
            {
                if(fabs(1-p[i][j][0]) < eps)
                    continue;
                if(i == n && j == m)
                    continue;
                if(j == m)
                    dp[i][j] += (dp[i+1][j]*p[i][j][2] + 2)/(1.0-p[i][j][0]);
                else if(i == n)
                    dp[i][j] += (dp[i][j+1]*p[i][j][1] + 2)/(1.0-p[i][j][0]);
                else
                    dp[i][j] += (dp[i+1][j]*p[i][j][2] + dp[i][j+1]*p[i][j][1] + 2)/(1.0-p[i][j][0]);
            }
        }
        printf("%.3lf\n",dp[1][1]);
    }
    return 0;
}



            




相關文章