POJ 2139 SIx Degrees of Cowvin Bacon 最短路 水題

　　　　　　　　　　　　　　　　　　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

USACO 2003 March Orange

 

 

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。

 

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int INF=0x3f3f3f3f;
const int MAXN=303;

int dp[MAXN][MAXN];
int tmp[MAXN];

void init(int N)
{
    for(int i=1;i<=N;i++)
    {
        for(int j=1;j<=N;j++)
        {
            if(i==j)
                dp[i][j]=0;
            else
                dp[i][j]=INF;
        }
    }
}

void floyd(int N)
{
    for(int k=1;k<=N;k++)
        for(int i=1;i<=N;i++)
            for(int j=1;j<=N;j++)
                dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
}

void solve(int N)
{
    int minc=INF;
    for(int i=1;i<=N;i++)
    {
        int sum=0;
        for(int j=1;j<=N;j++)
            sum+=dp[i][j];
        if(sum<minc)
            minc=sum;
    }
    printf("%d
",minc*100/(N-1));
}

int main()
{
    int N,M;
    while(~scanf("%d%d",&N,&M))
    {
        init(N);
        for(int i=0;i<M;i++)
        {
            int u;
            scanf("%d",&u);
            for(int j=1;j<=u;j++)
                scanf("%d",&tmp[j]);
            for(int j=1;j<=u;j++)
            {
                for(int k=1;k<=u;k++)
                {
                    if(j==k)
                        continue;
                    dp[tmp[j]][tmp[k]]=1;
                }
            }
        }
        floyd(N);

        solve(N);
    }
    return 0;
}