Six Degrees of Cowvin Bacon poj 2139

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 23 1 2 32 3 4

Sample Output

100

n头奶牛,m部电影,每部电影告诉你是哪几头奶牛.如果两头奶牛在一起出演就相互度=1,如果两头奶牛没有一起出演过,但是同时跟第三头奶牛出演过,那么这两头奶牛的相互度就等于他们与第三头奶牛的相互度之和。要求输出牛的最小平均分开度的100倍。

数据量又小，所以用Floyd

#include <cstdio>#include <cstring>const int maxn = 300 + 5;const int inf = 1<<29;int n, m;int f[maxn][maxn];int a[maxn];void floyd(){    int i, j, k;    for(k=1; k<=n; ++k)        for(i=1; i<=n; ++i)            for(j=1; j<=n; ++j)                if(f[i][j]> f[i][k] + f[k][j])                    f[i][j] = f[i][k] + f[k][j];}int main(){    int i, j, k, ans, ret;    while(~scanf("%d%d",&n, &m)) {        for(i=0; i<=n; ++i) {            for(j=0; j<=n; ++j) f[i][j] = inf;            f[i][i] = 0;        }        while(m--) {            scanf("%d", &k);            for(i=0; i<k; ++i) {                scanf("%d", &a[i]);                for(j=0; j<i; ++j)                    f[a[i]][a[j]] = f[a[j]][a[i]] = 1;            }        }        floyd();        ans = inf;        for(i=1; i<=n; ++i) {            ret = 0;            for(j=1; j<=n; ++j)                ret += f[i][j];            if(ans > ret) ans = ret;        }        printf("%d\n", ans*100/(n-1));    }    return 0;}