uva247电话圈（floyd）

简单的板子题，就是字符串操作有点难啊啊啊，

要熟悉下离散图论里连通图（有向图）里的可达性矩阵Warshall算法，（有向图的传递闭包）

如果d[i][j]&&d[j][i]都联通，则双方可达，

此题就用一个map和string数组存储输入的字符串，flag数组标记；

代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<map>
typedef long long ll;
const int maxn=1010;
const int inf=0x3f3f3f;
using namespace std;
int d[maxn][maxn];
int flag[maxn];
map<string,int> mp;
int n,m;
string name[maxn];
int sum;
string s1,s2;
void in()
{
    mp.clear();
    memset(flag,0,sizeof(flag));
    memset(d,0,sizeof(d));
    sum=0;
}
int x,y;
void enter()
{
    for(int i=1; i<=m; i++)
    {
      cin>>s1>>s2;
      if(mp.count(s1))
      {
          //cout<<"if"<<":1"<<endl;
          x=mp[s1];

      }
      else
      {
          //cout<<"else"<<":1"<<endl;
          mp[s1]=++sum;
          name[sum]=s1;
          x=sum;

      }
      if(mp.count(s2))
      {
          //cout<<"if"<<":2"<<endl;
          y=mp[s2];
      }
      else
      {
          //cout<<"else"<<":2"<<endl;
          mp[s2]=++sum;
          name[sum]=s2;
          y=sum;
      }
      //cout<<x<<" "<<y<<endl;
      d[x][y]=1;
    }

}
void floyd()
{
    for(int k=1; k<=m; k++)
        for(int i=1; i<=m; i++)
          for(int j=1; j<=m; j++)
           if(d[i][j]<inf&&d[j][k]<inf)
            d[i][j]=d[i][j]||(d[k][j]&&d[i][k]);
}
int main()
{
    int t=1;
   while(cin>>n>>m&&n+m)
   {
    if(t!=1) printf("
");
    in();
    enter();
    //for(map<string,int>::iterator it=mp.begin();it!=mp.end();it++)
        //cout<<it->first<<" "<<it->second<<endl;
    floyd();
    printf("Calling circles for data set %d:
",t++);
    for(int i=1; i<=n; i++)
        {
            if(flag[i]) continue;
            cout<<name[i];
            for(int j=i+1; j<=n; j++)
            {
                if(flag[j]) continue;
                if(d[i][j]&&d[j][i])
                {
                    flag[j]=1;
                    cout << ", " <<name[j];
                }
            }
            cout<<endl;
        }
   }
}