poj 2139 Six Degrees of Cowvin Bacon 最短路bellman 多源最短路径 （一次AC）

Six Degrees of Cowvin Bacon

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 3982Accepted: 1870

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

分析：bellman的模板题目，稍微有点价值的就是多源最短路径的处理

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define inf  0x3f3f3f3f
int dist[305][305],a[305];
int n,m,num,cnt;
struct Edge{
   int f,t,c;
}edge[1005];
void init()
{
     memset(dist,0x3f,sizeof(dist));
     for(int i=1;i<=1003;i++)
            edge[i].c=inf;
     cnt=0;
     for(int i=1;i<=n;i++)
        dist[i][i]=0;     //注意dist[i][i]要初始化为0
}
void bellman()
{
     for(int s=1;s<=n;s++)   //多源最短路径的处理，在模板中单个起点的while循环外嵌套一层不同起点的循环,同时dist改为二维
     {
        while(1)
        {
          bool flag=false;
          for(int j=1;j<=cnt;j++)
            {
                int from=edge[j].f,to=edge[j].t,cost=edge[j].c;
                if(dist[s][to]>dist[s][from]+cost)
                   {
                       dist[s][to]=dist[s][from]+cost;
                       flag=true;
                   }
            }
          if(!flag)
           break;
        }
     }
     int minn=inf,w;
     for(int i=1;i<=n;i++)
        {
           w=0;
           for(int j=1;j<=n;j++)
              w+=dist[i][j];
           if(w<minn)
              minn=w;
        }
     printf("%d\n",(minn*100)/(n-1));
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        init();
        for(int k=1;k<=m;k++)
        {
            scanf("%d",&num);
            for(int j=1;j<=num;j++)
                scanf("%d",&a[j]);
            for(int i=1;i<=num-1;i++)
                for(int j=i+1;j<=num;j++)
                    {
                        cnt++;
                        edge[cnt].f=a[i];
                        edge[cnt].t=a[j];
                        edge[cnt].c=1;
                        cnt++;
                        edge[cnt].f=a[j];
                        edge[cnt].t=a[i];
                        edge[cnt].c=1;
                    }
        }
        bellman();
    }
    return 0;
}