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 20.00 0.50 0.50 0.50 0.00 0.500.50 0.50 0.00 1.00 0.00 0.00
Sample Output
6.000
题意：有R*C个格子，一个家伙要从（1,1）走到（n,m） 每次只有三次方向，分别是不动，向下，向右，告诉你这三个方向的概率，以及每走一步需要耗费两个能量，问你走到终点所需要耗费能量的数学期望：
dp[i][j]代表 i,j到n,m需要的额能量
dp[i][j]=p[i][j][0]*dp[i][j]+dp[i][j+1]*p[i[[j][1]+dp[i+1][j]*p[i][j][2]+2 ;
变形就是 dp[i][j]=(dp[i][j+1]*p[i[[j][1]+dp[i+1][j]*p[i][j][2]+2)/(1-p[i][j][0]);
注意 p[i][j[0]==1 的情况
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b) for(i = a; i < b; i++)#define frer(i,a,b) for(i = a; i > =b;i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define ssf(n)scanf("%s", n)#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define bugpf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 1005double dp[N][N];// dp[i][j] 代表 i j 这个点需要的能量double p[N][N][3];int n,m;int main(){int i,j,k;while(~sff(n,m)){fre(i,1,n+1)fre(j,1,m+1)fre(k,0,3)scanf("%lf",&p[i][j][k]); dp[n][m]=0;for(i=n;i>=1;i–) for(j=m;j>=1;j–){if(i==n&&j==m) continue;if(fabs(p[i][j][0]-1)<eps) continue;dp[i][j]=(dp[i][j+1]*p[i][j][1]+dp[i+1][j]*p[i][j][2]+2)/(1-p[i][j][0]);}pf("%.3f\n",dp[1][1]);} return 0;}
，怪天怪地，我都不会怪你，你有选择幸福的权利…
