Probability DP: Flying chess hdu4405

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 has six faces with equal probability to face up and the numbers on the faces is 1, 2,3,4,5,6). When Hzz was at grid I and the dice number is x, he would moves to grid i+x. Hzz finishes the game when I+x is E Qual to or greater than N.

There is also M flight lines on the chess map. The i-th flight line can help Hzz fly from a grid Xi to Yi (0

#include <cstdio>
#include <cstring>
using namespace std;
int n,m,t1,t2, fly[100009];
Double dp[100009];
int main ()
{
    while (scanf ("%d%d", &n,&m) && (n| | m))
    {
        memset (fly,-1,sizeof fly);
        memset (dp,0,sizeof dp);
        while (m--)
        {
            int t2,t1;
            scanf ("%d%d", &t1,&t2);
            fly[t1]=t2;
        }
        for (int i=n-1;i>=0;i--)
        {
            if (fly[i]!=-1)
                Dp[i]=dp[fly[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;
}

Ideas:
1, if it is not possible to fly directly from a position can start to six different positions, the probability of the sum to the position, due to six possible equalization of opportunities, so the result is divided by 6 and plus one (no matter where you go will be the first dice;
2, if you can fly directly to the probability of flying to the location;