HDU 3853 LOOPS【期望DP】

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.

题解

联赛前日常水。。。

得

$$f_{i,j}=p1*f_{i+1,j}+p2*f_{i,j+1}+p0*f_{i,j}+2\\ f_{i,j}(1-p0)=p1*f_{i+1,j}+p2*f_{i,j+1}+2\\ f_{i,j}=\frac{p1*f_{i+1,j}+p2*f_{i,j+1}+2}{1-p0}$$
然后就直接推就好了。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define maxn 1006
using namespace std;
int n,m;
double p[maxn][maxn][3],f[maxn][maxn];
int main(){
    freopen("loops.in","r",stdin);
    freopen("loops.out","w",stdout);
    while(scanf("%d%d",&n,&m)==2){
        memset(f,0,sizeof(f));
        for(int i=1;i<=n;i++)
         for(int j=1;j<=m;j++)scanf("%lf%lf%lf",&p[i][j][0],&p[i][j][1],&p[i][j][2]);
        for(int i=n;i>=1;i--)
         for(int j=m;j>=1;j--) if(i<n||j<m){
            if(abs(p[i][j][0]-1)<1e-7)continue;
            f[i][j]=(p[i][j][2]*f[i+1][j]+p[i][j][1]*f[i][j+1]+2)/(1-p[i][j][0]);
         }
        printf("%.3lf\n",f[1][1]);
    }
    return 0;
}