POJ-2139 Six Degrees of Cowvin Bacon

题目传送门

Six Degrees of Cowvin Bacon

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 6179 Accepted: 2893

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

题意：

牛跟自己的分离度是0，如果两牛合作分离度则为1，如果两牛同时和第三头牛合作分离度为2.求一头牛到其他牛最小的平均分离度，即求最短路。

我个人比较喜欢用Dijkstra求最短路，虽然这个题目用floyd也可以，也更加方便，不过我还是用Dijkstra敲的代码。

基础最短路问题

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
using namespace std;
typedef pair<int,int>P;
const int maxn=500;
const int INF=0x3f3f3f3f;
struct Edge{
int to,len;
Edge(int to_=0,int len_=0)
{
to=to_;
len=len_;
}
};
vector<Edge>G[maxn];
int dis[maxn];
bool vis[maxn];
int n,m;
void init()
{
for(int i=1;i<=n;i++)
G[i].clear();
}
void Add(int from,int to,int len)
{
G[from].push_back(Edge(to,len));
G[to].push_back(Edge(from,len));
}
void Dijkstra(int s)
{
fill(dis,dis+n+1,INF);
dis[s]=0;
memset(vis,false,sizeof(vis));
priority_queue<P,vector<P>,greater<P> >que;
int u,v,i,len;
Edge e;
P p;
que.push(P(0,s));
while(!que.empty())
{
p=que.top();
que.pop();
u=p.second;
if(vis[u])
continue;
vis[u]=true;
len=G[u].size();
for(int i=0;i<len;i++)
{
e=G[u][i];
v=e.to;
if(!vis[v]&&dis[v]>dis[u]+e.len)
{
dis[v]=dis[u]+e.len;
que.push(P(dis[v],v));
}
}
}
}
int main()
{
while(scanf("%d%d",&n,&m)==2)
{
init();
for(int i=0;i<m;i++)
{
int k,ans[maxn],tot;
scanf("%d",&k);
tot=k;
while(k--)
{
scanf("%d",&ans[k]);
}
for(int i=0;i<tot;i++)
{
for(int j=0;j<tot;j++)
{
if(i==j)
continue;
Add(ans[i],ans[j],1);
}
}
}
int Min=INF;
for(int i=1;i<=n;i++)
{
Dijkstra(i);
int len_sum=0;
for(int i=1;i<=n;i++)
len_sum+=dis[i];
if(Min>len_sum)
Min=len_sum;
}
printf("%d\n",Min*100/(n-1));
}
}