HDU4405 Aeroplane chess

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2780    Accepted Submission(s): 1779

Problem Description

Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.

There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0
Please help Hzz calculate the expected dice throwing times to finish the game.

 

Input

There are multiple test cases. 
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi The input end with N=0, M=0. 

 

Output

For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.

 

Sample Input

2 0 8 3 2 4 4 5 7 8 0 0

 

Sample Output

1.1667 2.3441

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

题意：有一种飞行棋，每次摇骰子向前走，如果某个点有飞机，那就可以从那个点飞到另外一个点，如果到了一个点还有飞机就能继续向前飞行。问从起点飞到终点的期望是多少。

分析：概率DP求期望问题，http://kicd.blog.163.com/blog/static/126961911200910168335852/   这篇讲的不错

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
double dp[100009];
int f[100009];
int a,b;

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        if(n+m==0) break;

        memset(dp,0,sizeof dp);
        memset(f,-1,sizeof f);

        for(int i=0;i=0;i--)
        {
            if(f[i]!=-1)
                dp[i]=dp[f[i]];
            else
            {
                for(int j=1;j<=6;j++)
                    dp[i]+=dp[i+j];
                dp[i]=dp[i]/6.0+1;
            }
        }

        printf("%.4f\n",dp[0]);

    }
    return 0;
}