poj 2139 Six Degrees of Cowvin Bacon 最短路

Six Degrees of Cowvin Bacon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2213		Accepted: 1034

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

#include 
#include 
#include 
#include 
using namespace std;

const int INF=10000000;

int map[333][333];
int dis[333];
int vis[333];
int n,m;

void dijkstra(int s)
{
    int i,j,k,mn;
    for(i=1;i<=n;i++)
        dis[i]=map[s][i],vis[i]=0;
    vis[s]=1;
    for(i=1;i<=n;i++)
    {
        k=s;
        mn=INF;
        for(j=1;j<=n;j++)
            if(!vis[j]&&dis[j]>n>>m;
    for(i=1;i<=n;i++)
         for(j=1;j<=n;j++)
            if(i==j)
                map[i][j]=0;
            else
                map[i][j]=INF;
    for(i=0;i


发现自己偏爱dijkstra。。