Six Degrees of Cowvin Bacon（最短路）

原题链接
Six Degrees of Cowvin Bacon
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4687 Accepted: 2233
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

//http://poj.org/problem?id=2139#include <algorithm>#include <iostream>#include <utility>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>using namespace std;typedef long long ll;const int MOD = int(1e9) + 7;//int MOD = 99990001;const int INF = 0x3f3f3f3f;const ll INFF = (~(0ULL)>>1);const double EPS = 1e-9;const double OO = 1e20;const double PI = acos(-1.0); //M_PI;const int fx[] = {-1, 1, 0, 0};const int fy[] = {0, 0, -1, 1};const int maxn=300 + 20;int d[maxn][maxn];int x[maxn];int n,m;void floyd_warshall(){        for(int i=0;i<n;i++){                for(int j=0;j<n;j++){                        for(int k=0;k<n;k++){                                d[i][j]=min(d[i][j],d[i][k] + d[k][j]);                        }                }        }}int main(){        cin >> n >> m;        for(int i=0;i<maxn;i++)                for(int j=0;j<maxn;j++)                        d[i][j]=INF;        for(int i=0;i<n;i++){                d[i][i]=0;        }        while(m--){                int c;cin >> c;                for(int i=0;i<c;i++){                        cin >> x[i];                        x[i]--;                }                for(int i=0;i<c;i++){                        for(int j=i+1;j<c;j++){                                d[x[i]][x[j]] = 1;d[x[j]][x[i]] = 1;                        }                }        }        floyd_warshall();        int res=INF;        for(int i=0;i<n;i++){                int sum=0;                for(int j=0;j<n;j++)                        sum += d[i][j];                res=min(res,sum);        }        cout << res*100/(n-1) << endl;        return 0;}//上面的代码迷之WA,而下面的代码却AC了，始终想不明白原因/*#include <iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int n,m,cow[305][305],mo[305];const int MAX=100000;int main(){    cin>>n>>m;    for(int i=0;i<=n;i++)    {         fill(cow[i],cow[i]+n+1,MAX);         cow[i][i]=0;    }    while(m--)    {        int num;        scanf("%d",&num);        for(int i=0;i<num;i++)            scanf("%d",&mo[i]);        for(int i=0;i<num;i++)            for(int j=i+1;j<num;j++)            {                 cow[mo[i]-1][mo[j]-1]=1;                 cow[mo[j]-1][mo[i]-1]=1;            }    }    for(int k=0;k<n;k++)        for(int i=0;i<n;i++)        for(int j=0;j<n;j++)        cow[i][j]=min(cow[i][j],cow[i][k]+cow[k][j]);        int msum=1000000;    for(int i=0;i<n;i++)    {        int sum=0;        for(int j=0;j<n;j++)            sum+=cow[i][j];        if(sum<msum)            msum=sum;    }      printf("%d\n",msum*100/(n-1));    return 0;}*/