HDU_4405_AeroplaneChess

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2390    Accepted Submission(s): 1542

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

 

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zhoujiaqi2010

飞行棋的期望问题

每个位置的期望如果不能飞就要后面6种情况的期望

能飞要判断下是不是连续飞

#include 
#include 
using namespace std;

const int M=100010;
const int MM=1005;
double ex[M];
int fl[MM][2];

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)
            break;
        for(int i=1;i<=m;i++)
            scanf("%d%d",&fl[i][0],&fl[i][1]);
        ex[n]=ex[n+1]=ex[n+2]=ex[n+3]=ex[n+4]=ex[n+5]=0;//初始化的状态要足够多，并且符合题意
        for(int i=n-1;i>=0;i--)
        {
            int f=0,p;
            for(int j=1;j<=m;j++)//是否能飞
                if(fl[j][0]==i)
                {
                    f=1;
                    p=j;
                }
            if(f)              //是否能飞
            {
                ex[i]=ex[fl[p][1]]+1;
                int ff=0;
                for(int j=1;j<=m;j++)          //是否连续飞
                    if(fl[j][0]==fl[p][1])
                        ff=1;
                ex[i]-=1;
            }
            else
            ex[i]=1.0+(ex[i+1]+ex[i+2]+ex[i+3]+ex[i+4]+ex[i+5]+ex[i+6])/6.0;
        }
        printf("%.4lf\n",ex[0]);
    }
    return 0;
}