HDU 4405( Aeroplane chess)概率DP

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

题意：HZZ（胡智障？）下棋（编号0~N 掷骰子决定前进步数 ）

其中有一些便捷通道 可以从i直达j 询问到达N时掷骰子次数的期望

继续概率dp练习

到达便捷通道时dp[j]=dp[i]，其它点dp[i]=dp[j]/6.0

按照这个规律得到状态转移方程

#include<bits/stdc++.h>
#define LL long long
#define maxn 101000
#define eps 1e-8
using namespace std;
double dp[maxn];
int route[maxn];
int main()
{
    //ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int n,m;
    while(~scanf("%d%d",&n,&m)&&n)
    {
        memset(dp,0,sizeof(dp));
        memset(route,-1,sizeof(route));
        while(m--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            route[a]=b;
        }
        for(int i=n-1;i>=0;i--)
        {
            if(route[i]==-1)
            {
                for(int j=1;j<=6;j++)
                {
                    dp[i]+=dp[i+j]/6;
                }
                dp[i]++;
            }
            else
            {
                dp[i]=dp[route[i]];
            }
        }
        printf("%.4f\n",dp[0]);
    }
    return 0;
}

继续加油！