hdu4405Aeroplane chess(概率与期望dp)

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4239    Accepted Submission(s):
2674

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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/*
本题是扔色子，其实概率也就是1/6，但是本题还有一个不同的地方，那就是可以飞行
比如从a飞到b，而不用扔色子，这些条件都会影响我们做题的方法
首先定义状态dp[i]表示处于i位置时所扔色子次数的期望，可知dp[n] = 0。
然后找找转移方程，那根据题意可知，i位置要么是可以飞行的，要么就不是
所以当i点可以飞行时，dp[i] = dp[fly[i]]，其中fly[i]代表i飞行到的另一个点
否则，依据期望的定义，dp[i] = 1/6*dp[i+j] + 1，其中j = 1，2...6，代表扔色子可能的点数
最后别忘了加1，代表本次需要扔一次色子。最后，规划方向依然是从后往前，答案就是dp[0]。
*/
#include<iostream>
#include<cstdio>
#include<cstring>

#define N 100005

using namespace std;
double dp[N];
int fly[N];

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m) && (n||m))
    {
        for(int i=;i<=n;i++)
        {
            dp[i]=;fly[i]=-;
        }
        int a,b;
        for(int i=;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            fly[a]=b;
        }
        for(int i=n-;i>=;i--)
        {
            if(fly[i]!=-) dp[i]=dp[fly[i]];
            else
            {
                for(int j=;j<=;j++)
                {
                    if(i+j>n) dp[i]+=1.0/*dp[n];
                    else dp[i]+=1.0/*dp[i+j];
                }++dp[i];
            }
        }
        printf("%.4f\n",dp[]);
    }
    return ;
}