HDU 4405 Aeroplane chess（概率dp）

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3119    Accepted Submission(s): 1993

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<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.

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<Yi≤N).  
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 08 32 44 57 80 0

 

Sample Output

1.16672.3441

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

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zhoujiaqi2010

题目大意：

    玩飞行棋，有一个骰子可以走1～6步，还有一些航线可以直接从一个格子移动到前面的一个格子。求走到终点的步数期望。

解题思路：

    从终点到起点递推，如果一个格子有航线，那么它的期望就是航线所连接的前面的格子的期望，否则它的期望就是前面6步格子期望除以6再加一。

附AC代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;const int maxn=100000+5;int path[maxn],n,m;double dp[maxn];int main(){    while(~scanf("%d%d",&n,&m)&&(n||m))    {        memset(path,-1,sizeof path);//-1表示没有航线        for(int i=0;i<m;++i)        {            int from;            scanf("%d",&from);            scanf("%d",&path[from]);        }        memset(dp,0,sizeof dp);//初始期望为0        for(int i=n-1;i>=0;--i)        {            if(path[i]!=-1)//有航线                dp[i]=dp[path[i]];            else            {                for(int j=1;j<=6;++j)                    dp[i]+=dp[i+j];                dp[i]=dp[i]/6+1;            }        }        printf("%.4f\n",dp[0]);    }        return 0;}