POJ2139 Six Degrees of Cowvin Bacon 【Floyd】

Six Degrees of Cowvin Bacon

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3131 Accepted: 1455

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 23 1 2 32 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 <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>#include <queue>using namespace std;typedef long long LL;#define maxn 302#define inf 0x3f3f3f3fint G[maxn][maxn], N, M;int sta[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(G[i][k] + G[k][j] < G[i][j])                    G[i][j] = G[i][k] + G[k][j];  }int main() {    // freopen("stdin.txt", "r", stdin);    int i, j, num, ans = inf;    memset(G, 0x3f, sizeof(G));    scanf("%d%d", &N, &M);    for(i = 1; i <= N; ++i)        G[i][0] = G[i][i] = 0;    while(M--) {        scanf("%d", &num);        for(i = 0; i < num; ++i) {            scanf("%d", sta + i);            for(j = 0; j < i; ++j)                G[sta[i]][sta[j]] = G[sta[j]][sta[i]] = 1;        }    }    Floyd();    for(i = 1; i <= N; ++i) {        for(j = 1; j <= N; ++j)            G[i][0] += G[i][j];        if(ans > G[i][0]) ans = G[i][0];     }    printf("%d\n", ans * 100 / (N - 1));    return 0;}