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 2 3 1 2 3 2 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 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
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 else
22 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<minc)
44 minc=sum;
45 }
46 printf("%d\n",minc*100/(N-1));
47 }
48
49 int main()
50 {
51 int N,M;
52 while(~scanf("%d%d",&N,&M))
53 {
54 init(N);
55 for(int i=0;i<M;i++)
56 {
57 int u;
58 scanf("%d",&u);
59 for(int j=1;j<=u;j++)
60 scanf("%d",&tmp[j]);
61 for(int j=1;j<=u;j++)
62 {
63 for(int k=1;k<=u;k++)
64 {
65 if(j==k)
66 continue;
67 dp[tmp[j]][tmp[k]]=1;
68 }
69 }
70 }
71 floyd(N);
72
73 solve(N);
74 }
75 return 0;
76 }