HDU 4403（Aeroplane chess ，求期望，概率DP）

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

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

Recommend

zhoujiaqi2010

 

題目大意：

飛行棋。給一組數據 N，M ，N代表有N+1（一維，0->N）個格子，你的起始點是在0號位置，M代表你有M條航班，接下來會有M行，每行兩個整數X,Y，表示在位置X和位置Y有一條航班，可以直接從X飛到Y，投擲一枚骰子，投擲多少就能走多少步，如遇到航班，則按照航班走，航班可以連續，每個航班的起始點不同。輸出投擲色子次數的期望。

解題思路：

dp [ n ]=0，dp [ i ]=sum( dp [i+j] ) +1, j 從1累加到6，因爲期望代表的是步數，所以每次加 1 步，當遇到航班 (x,y)時，則記dp [ x ] = dp [ y ] ，則結果等於dp [0].

代碼：

#include<iostream>
#include<cstdio>

using namespace std;

const int maxN=100010;

int main(){
    int a,b,n,m,visited[maxN];
    double dp[maxN];
    while(~scanf("%d%d",&n,&m)&&(n||m)){
        for(int i=0;i<n;i++){
            visited[i]=-1;
        }
         for(int i=0;i<=n+5;i++){
            dp[i]=0;
        }
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            visited[a]=b;
        }
        for(int i=n-1;i>=0;i--){
            if(visited[i]==-1){
                for(int j=1;j<=6;j++){
                    dp[i]=dp[i+j]/6.0+dp[i];
                }
                dp[i]+=1;
            }
            else dp[i]=dp[visited[i]];

        }
        printf("%.4lf\n",dp[0]);
    }
    return 0;
}