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

题目大意：

飞行棋。给一组数据 N，M ，N代表有N+1（一维，0->N）个格子，你的起始点是在0号位置，M代表你有M条航班，接下来会有M行，每行两个整数X,Y，表示在位置X和位置Y有一条航班，可以直接从X飞到Y，投掷一枚骰子，投掷多少就能走多少步，如遇到航班，则按照航班走，航班可以连续，每个航班的起始点不同。输出投掷色子次数的期望。

解题思路：

dp [ n ]=0，dp [ i ]=sum( dp [i+j] ) +1, j 从1累加到6，因为期望代表的是步数，所以每次加 1 步，当遇到航班 (x,y)时，则记dp [ x ] = dp [ y ] ，则结果等于dp [0].

代码：

#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;
}