POJ 2139 Six Degrees of Cowvin Bacon （floyd)

就是用floyd求一个点点最短距离然后求个平均值

 

 

题目：

Six Degrees of Cowvin Bacon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2359		Accepted: 1116

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

 

代码：

#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
using namespace std;
#define INF 0x7ffffff
int G[305][305];
int N,M;
void input()
{
    scanf("%d%d",&N,&M);

    for(int i=0;i<=N;i++)
    {
        for(int j=0;j<=N;j++)
        {
            G[i][j] = INF;
        }
    }
    vector<int> v;
    for(int i=0;i<M;i++)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            int t; scanf("%d",&t);
            v.push_back(t);
        }
        int len = v.size();
        for(int p=0;p<len;p++)
        {
            for(int q=p+1;q<len;q++)
            {
                G[v[p]][v[q]] = G[v[q]][v[p]] = 1;
            }
        }
        v.clear();
    }
}
void floyd()
{

    for(int k=1;k<=N;k++)
    {
        for(int i=1;i<=N;i++)
        {
            for(int j = 1;j<=N;j++)
            {
                if(i!=j)
                G[i][j] = min( G[i][j] , G[i][k] + G[k][j]);
            }
        }
    }
}

void print()
{
    for(int i=1;i<=N;i++)
    {
        for(int j=1;j<=N;j++)
        {
            cout<<G[i][j]<<' ';
        }
        cout<<endl;
    }

}
int main()
{

    input();
   // print();
    floyd();

    double ans = 100000000;
    for(int i=1;i<=N;i++)
    {
            int deg = 0 , cnt =0;
            for(int j=1;j<=N;j++)
            {
                if( i == j)continue;

                if( G[i][j] != INF)
                {
                    deg += G[i][j];
                    cnt++;
                }
            }
            if( ans > deg*1.0/cnt)
            {
                ans = deg*1.0/cnt;
            }
    }
    cout<<int(ans*100)<<endl;
    return 0;
}