hdu 4405 Aeroplane chess

Aeroplane chess

http://acm.hdu.edu.cn/showproblem.php?pid=4405

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4090    Accepted Submission(s): 2607

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 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

 

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题意：

0——n的格子上玩飞行棋，每次抛色子，点数若为j，下一步到达当前位置+j处

格子中有可以飞的点，可以从i直接飞到j

>=n都可以视作终点

抛色子抛出的点的概率相同，问到终点期望抛几次

 

dp[i]表示i到终点的期望次数

dp[n]=0,ans=dp[0]

若i可以飞到j，dp[i]=dp[j]

否则 dp[i]= （Σ dp[i+k] / 6 ）+1, 1<=k<=6  

 

#include<cstdio>
#include<cstring>
#define N 100001
using namespace std;
int fly[N];
double dp[N];
int main()
{
    int n,m,x,y;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(!n) return 0;
        memset(fly,0,sizeof(fly));
        memset(dp,0,sizeof(dp));
        while(m--)
        {
            scanf("%d%d",&x,&y);
            fly[x]=y;
        }
        for(int i=n-1;i>=0;i--)
        {
            if(fly[i]) dp[i]=dp[fly[i]];
            else
            {
                for(int j=1;j<=6;j++)
                 if(i+j<n) dp[i]+=dp[i+j];
                dp[i]=dp[i]/6.0+1;
            }
        }
        printf("%.4lf
",dp[0]);
    }
}