Six Degrees of Cowvin Bacon
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3602 | Accepted: 1675 |
Description
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 <= N <= 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 <= M <= 10000) movies and it is guaranteed that some relationship path exists between every pair of cows.
Input
* Line 1: Two space-separated integers: N and M * Lines 2..M+1: Each input line contains a set of two or more space-separated integers that describes the cows appearing in a single movie. The first integer is the number of cows participating in the described movie, (e.g., Mi); the subsequent Mi integers tell which cows were.
Output
* Line 1: A single integer that is 100 times the shortest mean degree of separation of any of the cows.
Sample Input
4 23 1 2 32 3 4
Sample Output
100
Hint
[Cow 3 has worked with all the other cows and thus has degrees of separation: 1, 1, and 1 -- a mean of 1.00 .]
Source
好吧,這道題題意我沒有看懂,最後是google了別人的題解的題意的。
題意:為了向一個理論“任何人可以通過最多5個人認識任何一個人”致敬,一群牛決定開拍電影,
有N頭牛,拍了M電影,每一部電影裡面有u個角色。
若2頭牛在同一部電影中出現,則他們就認識了,距離為1,若2部牛沒有在同一部電影裡面出現過,但是他們都認識第3頭牛,則他們的距離為2,一次類推。
要求:
找出一頭牛,他到其他所有牛的距離之和sum是最小的,輸出最小的平均距離,即sum/(N-1),
向下取整。
floyd就好啦。
1 #include2 #include 3 #include 4 5 using namespace std; 6 7 const int INF=0x3f3f3f3f; 8 const int MAXN=303; 9 10 int dp[MAXN][MAXN];11 int tmp[MAXN];12 13 void init(int N)14 {15 for(int i=1;i<=N;i++)16 {17 for(int j=1;j<=N;j++)18 {19 if(i==j)20 dp[i][j]=0;21 else22 dp[i][j]=INF;23 }24 }25 }26 27 void floyd(int N)28 {29 for(int k=1;k<=N;k++)30 for(int i=1;i<=N;i++)31 for(int j=1;j<=N;j++)32 dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);33 }34 35 void solve(int N)36 {37 int minc=INF;38 for(int i=1;i<=N;i++)39 {40 int sum=0;41 for(int j=1;j<=N;j++)42 sum+=dp[i][j];43 if(sum