9 人解决，9 人已尝试。
13 份提交通过，共有 27 份提交。
5.0 EMB 奖励。
单点时限: 1.0 sec
内存限制: 256 MB
The cows have been making movies lately, so they are ready to play a variant of the famous game “Six Degrees of Kevin Bacon”.
The game works like this: each cow is considered to be zero degrees of separation (degrees) away from herself. If two distinct cows have been in a movie together, each is considered to be one ‘degree’ away from the other. If a two cows have never worked together but have both worked with a third cow, they are considered to be two ‘degrees’ away from each other (counted as: one degree to the cow they’ve worked with and one more to the other cow). This scales to the general case.
The $N$ $(2 \leq N \leq 300)$ cows are interested in figuring out which cow has the smallest average degree of separation from all the other cows. excluding herself of course. The cows have made $M$ $(1 \leq M \leq 10000)$ movies and it is guaranteed that some relationship path exists between every pair of cows.
A single integer that is $100$ times the shortest mean degree of separation of any of the cows.
4 2 3 1 2 3 2 3 4
100
Cow 3 has worked with all the other cows and thus has degrees of separation: 1, 1, and 1 – a mean of 1.00 .
9 人解决，9 人已尝试。
13 份提交通过，共有 27 份提交。
5.0 EMB 奖励。