HDU-3853-期望/dp/坑

Posted 2020-11-02 zzq

LOOPS

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 7826    Accepted Submission(s): 3156

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

 

　　　　　　f[i][j]表示位于(i,j)距离目标的期望魔力值，可是对于朴素点如果p1=1的话那显然期望就是inf了因为会陷入死循环，特判一下改成0就过了。

　　　　

 1 #include<iostream>
 2 #include<cstring>
 3 #include<queue>
 4 #include<cstdio>
 5 #include<stack>
 6 #include<set>
 7 #include<map>
 8 #include<cmath>
 9 #include<ctime>
10 #include<time.h> 
11 #include<algorithm>
12 using namespace std;
13 #define mp make_pair
14 #define pb push_back
15 #define debug puts("debug")
16 #define LL long long 
17 #define pii pair<int,int>
18 #define eps 1e-10
19 
20 double f[1010][1010];
21 double p[1010][1010][3];
22 int main()
23 {
24     int n,m,i,j,k,t;
25     while(scanf("%d%d",&n,&m)==2){
26         memset(f,0,sizeof(f));
27         for(i=1;i<=n;++i){
28             for(j=1;j<=m;++j){
29                 for(k=0;k<3;++k)
30                   scanf("%lf",&p[i][j][k]);
31             }
32         }
33         f[n][m]=0;
34         for(i=n;i>=1;--i){
35             for(j=m;j>=1;--j){
36                 if(i==n&&j==m) continue;
37                 double p1=p[i][j][0],
38                        p2=p[i][j][1],
39                        p3=p[i][j][2];
40                 if(fabs(p1-1)<=eps) {
41                     f[i][j]=0;
42                     continue;
43                 }
44                 f[i][j]=(p2*f[i][j+1]+p3*f[i+1][j]+2)/((double)1-p1);
45             }
46         }
47         printf("%.3f\n",f[1][1]);
48     }
49     return 0; 
50 }