POJ 2139 Six Degrees of Cowvin Bacon （floyd）

Six Degrees of Cowvin Bacon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6744		Accepted: 3139

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

 

题意稍微有点难理解，就是如果两头牛出现在同一个电影里（同一行），他们的分离度是1，如果不在一行，就要通过间接方式，比方说AB在同一场，BC在同一场，那么AC分离度就是2，以此类推，A和A分离度是0。求一头和其他牛分离度最低的牛，输出这n-1个和的平均数，然后把结果乘100。（最好是先乘100再除，免得各种转换数据类型，因为题目要求输出int类型）。

思路就是floyd（因为最多300头牛）。

//#include
#include
#include
#include
#include
#include<string>
#include
#include
#include
using namespace std;
#define maxn 305
#define INF 99999999
typedef pair<int,int> pii;
vector e[maxn];
int d[maxn],vis[maxn],mp[maxn][maxn];
int n,m;
void init()
{
    for(int i=1; i)
    {
        for(int j=1; j)
        {
            if(i==j) mp[i][j]=0;
            else mp[i][j]=INF;
        }
    }
}

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


int main()
{
    init();
    scanf("%d %d",&n,&m);

    while(m--)
    {
        int t,a[maxn];
        scanf("%d",&t);
        for(int i=0; i)
            scanf("%d",&a[i]);
        for(int i=0; i)
            for(int j=i+1; j)
            {
                mp[a[i]][a[j]]=mp[a[j]][a[i]]=1;
            }
    }
    floyd();
    int ans=INF;
    for(int i=1; i<=n; i++)
    {
        int sum=0;
        for(int j=1; j<=n; j++)
        {
            sum+=mp[i][j];
        }
        ans=min(sum,ans);
    }
    int ass=ans*100/(n-1);
    printf("%d\n",ass);
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/youchandaisuki/p/8826228.html