传送门：http://poj.org/problem?id=2139
Six Degrees of Cowvin Bacon
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6709 Accepted: 3122
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 .]
中文题意：
数学课上，WNJXYK忽然发现人缘也是可以被量化的，我们用一个人到其他所有人的平均距离来量化计算。
在这里定义人与人的距离：
1.自己与自己的距离为0
2.如果A和B属于同一个小团体，那么他们之间的距离为1
3.如果A与B属于一个小团体，B与C属于一个小团体，且A与C不同属于任何一个小团体，那么A与C的距离为2（A联系C，经过B、C两个人）
4.以此类推
班里有N个人 (2 <= N <= 300)，共有M对小团体关系(1 <= M <= 10000)。现在，给你所有班级中小团体的信息，问你班里人缘最好的人到其他人的平均距离。（你需要输出平均距离的100倍）
Input
第一行输入两个证书N，M 接下来的M+1行。
每行开头一个整数K表示本小团体大小，然后接下来K个整数表示小团体内学生编号。
Output
输出一行：最小的平均距离*100（程序中请不要使用Float和Double计算，可能会判Wa）
Sample Input
4 2
3 1 2 3
2 3 4
Sample Output
100
解题心得：
- 其实就是一个建图的问题，直接将同一个团体里面的人都添加一条边，距离是100，不建立其他的边，枚举每个点跑最短路就行了。
#include <stdio.h>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int maxn = 310;
int maps[maxn][maxn],Time[maxn],Min = 0x7f7f7f7f;
bool vis[maxn];
int n,m;
void init() {
int num[10010];
memset(maps,0,sizeof(maps));
for(int z=0;z<m;z++) {
int cnt;
scanf("%d",&cnt);
for(int i=0;i<cnt;i++) {
scanf("%d",&num[i]);
}
for(int i=0;i<cnt;i++)
for(int j=0;j<i;j++) {
maps[num[i]][num[j]] = maps[num[j]][num[i]] = 100;
}
}
}
void spfa(int s) {
memset(Time,0x7f,sizeof(Time));
queue <int> qu;
qu.push(s);
Time[s] = 0;
while(!qu.empty()) {
int now = qu.front(); qu.pop();
vis[now] = false;
for(int i=1;i<=n;i++) {
if(maps[now][i]) {
if(Time[i] > Time[now] + maps[now][i]) {
Time[i] = Time[now] + maps[now][i];
if(!vis[i]) {
qu.push(i);
vis[i] = true;
}
}
}
}
}
}
void get_sum_path() {
int sum = 0;
for(int i=1;i<=n;i++) {
sum += Time[i];
}
if(sum < Min)
Min = sum;
}
int main() {
scanf("%d%d",&n,&m);
init();
for(int i=1;i<=n;i++) {
spfa(i);
get_sum_path();
}
printf("%d",Min/(n-1));
return 0;
}