http://acm.hdu.edu.cn/showproblem.php?pid=3853
LOOPS
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2263 Accepted Submission(s):
911
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
等于1的时候就要continue;
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
double e[2000][2000];
struct node
{
double p1,p2,p3;
}a[2000][2000];
int main()
{
int r,c,i,j;
while(~scanf("%d%d",&r,&c))
{
memset(e,0,sizeof(e));
for(i=1;i<=r;i++)
for(j=1;j<=c;j++)
{
scanf("%lf%lf%lf",&a[i][j].p1,&a[i][j].p2,&a[i][j].p3);
}
e[r][c]=0;
for(i=r;i>=1;i--)
{
for(j=c;j>=1;j--)
{
if(i==r&&j==c)
continue;
if(a[i][j].p1==1)
continue;
e[i][j]=double(a[i][j].p2*e[i][j+1]+a[i][j].p3*e[i+1][j]+2)/double(1-a[i][j].p1);
}
}
printf("%.3lf\n",e[1][1]);
}
return 0;
}