HUD：4405-Aeroplane chess（期望飞行棋）

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

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

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解题心得：

1. 很经典的一个期望dp的题，必然事件是在第n-1个格子期望抛掷一次就结束游戏，所以从这个已知的点来推为止的期望，也就成了逆推期望。在可以飞行的点，前面的点的期望值就是其飞行到达点的期望值，然后顺其自然的就可以状态转移了。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
int to[maxn],n,m;
double dp[maxn];
void init()
{
    memset(dp,0,sizeof(dp));
    memset(to,-1,sizeof(to));
    while(m--)
    {
        int s,e;
        cin>>s>>e;
        to[s] = e;//记录飞行的到达的点
    }
}

void solve()
{
    for(int i=n-1;i>=0;i--)
    {
        if(to[i] != -1)
            dp[i] = dp[to[i]];
        else
        {
            for(int k=i+1;k<=i+6;k++)//抛出大于个格子数的点也是合法的，所以骰子每一面的可能性一直都是一样的
                dp[i] += dp[k]/6.0;
            dp[i] += 1.0;
        }
    }
    printf("%.4f\n",dp[0]);
}

int main()
{
    while(cin>>n>>m && n+m)
    {
        init();
        solve();
    }
    return 0;
}