HDU 3853 LOOPS（概率dp 求期望）

标签：dp   hdu   

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3853

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.

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

6.000

题意：

一个人 从一个为r行c列的矩形的左上角走到右下角！

每个点有一个如果有一定的概率分别传送到另一个点！

求她需要的期望值！

代码如下：

#include <cstdio>
#include <cstring>
double dp[1017][1017];
double a[1017][1017][4];
//dp[i][j] = dp[i][j]*a[i][j][1]+dp[i][j+1]*a[i][j][2]+dp[i+1][j]*a[i][j][3]+2;
//每一步需要消耗两点魔力值所以加2
//移项得：dp[i][j] = (dp[i][j+1]*a[i][j][2]+dp[i+1][j]*a[i][j][3]+2)/(1.0-a[i][j][1]);
int main()
{
    int r, c;
    while(~scanf("%d%d",&r,&c))
    {
        for(int i = 1; i <= r; i++)
        {
            for(int j = 1; j <= c; j++)
            {
                for(int k = 1; k <= 3; k++)
                {
                    scanf("%lf",&a[i][j][k]);
                }
            }
        }
        memset(dp, 0,sizeof(dp));
        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(a[i][j][1] == 1)
                {
                    continue;//一定在原地
                }
                dp[i][j] = (dp[i][j+1]*a[i][j][2]+dp[i+1][j]*a[i][j][3]+2)/(1.0-a[i][j][1]);
            }
        }
        printf("%.3lf\n",dp[1][1]);
    }
    return 0;
}

HDU 3853 LOOPS（概率dp 求期望）

标签：dp   hdu   

原文地址：http://blog.csdn.net/u012860063/article/details/45539689