問題：

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.?


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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?

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Output

A real number at 3 decimal places (round to), representing the expect magic power Homura need to escape from the LOOPS.?
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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

題意：有一個R*C的網格，每走一次消耗2點能量，現在要從起點 (1,1) 走到 (R , C),有三種走法：第一種原點不動；第二種向下走；第三種：向右走。三種走法的概率之和為1。求走到出口需要能量值的期望值。

思路：dp期望值的入門題，直接套數學期望值的公式即可，代碼有詳解。

代碼：

#include<stdio.h> #include<string.h> #include<math.h> #include<queue> #include<map> #include<stdlib.h> #include<iostream> #include<algorithm> #include<vector> #include<string> #define mem(a,b) memset(a,b,sizeof(a)) #define inf 0x3f3f3f3f #define ll long long #define PI acos(-1.0) #define eps 1e-8 #define mod 1000000007 #define lowbit(x) x&(-x) #define N 1010 //原點的概率，往右走一步的概率，往下走一步的概率分別用p1,p2,p3表示， //則期望等于dp[i][j]=dp[i][j]*p1+dp[i][j+1]*p2+dp[i+1][j]*p3+2(2為移動一次的消耗的能量)化簡為 //dp[i][j]=(dp[i][j+1]*p2+dp[i+1][j]*p3+2)/(1-p1); double dp[N][N]; double s[N][N][3];//表示原點，往右走一步，往下走一步 double a,b,c,d; int n,m; int main() {while(~scanf("%d%d",&n,&m)){mem(dp,0.0);for(int i=1; i<=n; i++)for(int j=1; j<=m; j++)for(int k=0; k<3; k++)scanf("%lf",&s[i][j][k]);//這道題比較簡單，不用再去算概率了for(int i=n; i>0; i--)for(int j=m; j>0; j--){a=1.0*s[i][j][0];//原點的概率b=1.0*dp[i][j+1]*s[i][j][1];//往右走一步的期望c=1.0*dp[i+1][j]*s[i][j][2];//往下走一步的期望if(i==n&&j==m)continue;if(1.0-a<eps)continue;//這一步如果1.0-a==0，是不能進行下一步的，分母不能為零dp[i][j]=(b+c+2.0)/(1.0-a);}printf("%.3f\n",dp[1][1]);} }

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