HDU 3853 LOOPS 概率DP入门

LOOPS

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 8453    Accepted Submission(s): 3397

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

概率DP入门，自己推一推公式就odk了。

不过想想队友给别的实验室的孩子出概率DP我就心惊肉跳233333

以及可能写三维的可读性更强。

 #include<bits/stdc++.h>
 using namespace std;
 #define mem(a,b) memset(a,b,sizeof(a))
 #define ll long long
 #define inf 1000000000
 #define maxn 1005
 #define maxm 100005
 #define eps 1e-10
 #define for0(i,n) for(int i=1;i<=(n);++i)
 #define for1(i,n) for(int i=1;i<=(n);++i)
 #define for2(i,x,y) for(int i=(x);i<=(y);++i)
 #define for3(i,x,y) for(int i=(x);i>=(y);--i)
 #define mod 1000000007
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>'') {if(ch=='-') f=-;ch=getchar();}
     while(ch>=''&&ch<='') {x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 double dp[maxn][maxn];
 double p1[maxn][maxn],p2[maxn][maxn],p3[maxn][maxn];
 int main()
 {
     int r,c;
     while(~scanf("%d%d",&r,&c))
     {
         for(int i=;i<=r;++i)
          for(int j=;j<=c;++j)
            scanf("%lf%lf%lf",&p1[i][j],&p2[i][j],&p3[i][j]);
         mem(dp,);
         for(int i=r;i>=;--i)
          for(int j=c;j>=;--j)
           {
               if(i==r&&j==c) continue;
               if(p1[i][j]==1.00) continue;
               dp[i][j]=(p2[i][j]*dp[i][j+]+p3[i][j]*dp[i+][j]+)/(-p1[i][j]);
           }
         printf("%.3lf\n",dp[][]);
     }
 }