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.
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.
* 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 .]
题意：牛牛们要拍大电影啦！拍同一部电影的牛距离为1，问你一头牛到其他所有牛距离的平均值的最小值是多少？
题解：裸的Floyd-Warshall算法，直接做就行了
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 310;
int d[maxn][maxn];
int g[maxn]; //记录同一个电影中的牛
int n, m;
void floyd()
{
for (int k = 0; k < n; ++k)
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
}
int main()
{
cin >> n >> m;
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
d[i][j] = (i == j) ? 0 : INF;
while (m--)
{
int v;
scanf("%d", &v);
for (int i = 0; i < v; ++i)
{
scanf("%d", &g[i]);
g[i]--;
}
for (int i = 0; i < v; ++i)
for (int j = i+1; j < v; ++j)
d[g[i]][g[j]] = d[g[j]][g[i]] = 1;
}
floyd();
int ans = INF;
for (int i = 0; i < n; ++i)
{
int sum = 0;
for (int j = 0; j < n; ++j)
sum += d[i][j];
ans = min(ans, sum);
}
cout << 100 * ans / (n - 1) << endl;
return 0;
}