【动态规划09】hdu3853 LOOPS（期望dp）

最新推荐文章于 2020-12-06 17:12:39 发布

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#dp #动态规划 #期望 #hdu

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本文介绍了一个基于概率转移的动态规划算法，用于计算魔法少女Homura从迷宫左上角到右下角所需的最小期望魔力。迷宫由R*C网格构成，每个格子包含一个传送门，传送门会按特定概率将角色传送到下方、右侧或原地。

题目描述

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.

总而言之就是一个r*c的矩阵，每次移动的cost是2，每次可以向右向下或者不动给你三种情况的概率，求从
(1,1)到(r,c)的最小期望膜力。
很容易想到设f[i][j]表示从(i,j)到(r,c)所需要的膜力期望。
f[i][j]一共可以到达f[i+1][j],f[i][j+1]和f[i][j]。
易得到递推式f[i][j]=(2+f[i][j])*mp[i][j][0]+(2+f[i][j+1]*mp[i][j][1]+(2+f[i+1][j])*mp[i][j][2]
移向得到f[i][j]= $2*(mp[i][j][0]+mp[i][j][1]+mp[i][j][2])+(2+f[i][j+1]*mp[i][j][1]+(2+f[i+1][j])*mp[i][j][2]\over{1-mp[i][j][0]}$
需要注意的是特殊判断分母不能为0

#include<bits/stdc++.h>
#define fer(i,j,n) for(int i=j;i<=n;i++)
#define far(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
const int maxn=1010;
const int INF=1e9+7;
using namespace std;
/*----------------------------------------------------------------------------*/
inline int read()
{
    char ls;int x=0,sng=1;
    for(;ls<'0'||ls>'9';ls=getchar())if(ls=='-')sng=-1;
    for(;ls>='0'&&ls<='9';ls=getchar())x=x*10+ls-'0';
    return x*sng;
}
/*----------------------------------------------------------------------------*/
double mp[maxn][maxn][3],f[maxn][maxn];
int n,m;
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        fer(i,1,n)
            fer(j,1,m)
                fer(k,0,2)
                    scanf("%lf",&mp[i][j][k]);
        f[n][m]=0;
        far(i,n,1)
            far(j,m,1)
            {
                if(i==n&&j==m)continue;
                if(mp[i][j][0]==1)continue;
                f[i][j]=(2+f[i][j+1]*mp[i][j][1]+f[i+1][j]*mp[i][j][2])/(1-mp[i][j][0]);  
            }
    printf("%.3lf\n",f[1][1]);
    }
}