Ural 1004 Sightseeing trip

Ural 1004 Sightseeing trip

第一次提交就WA了，而且在第一次测试点就WA了！瞅了半天，没看出来哪错……最后发现，"No solution"少了一个"."！无语无语……淡定淡定……
看别人用的是Floyed求最小环，确实很方便！学习了！
我用的是枚举起点终点+Dijkstra的方法。
以下是我的代码：

#include < iostream >
#include < string .h >
#define  maxn 107
#define  INF 20000007
using   namespace  std;
long  n,m,ans,num,g[maxn][maxn],tmp[maxn],path[maxn];
long  dijkstra( long  begin, long  end)
{
     long  min,minp,d[maxn];
     bool  used[maxn];
     for ( long  i = 1 ;i <= n;i ++ )
        d[i] = (i == begin ? 0 :INF);
    memset(used, false , sizeof (used));
    memset(tmp, 0 , sizeof (tmp));
    
     for ( long  i = 1 ;i <= n;i ++ )
    {
        min = INF;minp = 0 ;
         for ( long  j = 1 ;j <= n;j ++ )
             if ( ! used[j] && min > d[j])
            {
                min = d[j];minp = j;
            }
        used[minp] = true ;
         for ( long  j = 1 ;j <= n;j ++ )
             if ( ! used[j] && d[j] > min + g[minp][j])
            {
                d[j] = min + g[minp][j];
                tmp[j] = minp;
            }
    }
     return  d[end];
}
int  main()
{
     while (cin >> n >> m  &&  n !=- 1 )
    {
         for ( long  i = 1 ;i <= n;i ++ )
             for ( long  j = 1 ;j <= n;j ++ )
                g[i][j] = INF;
         for ( long  i = 1 ;i <= m;i ++ )
        {
             long  u,v,w;
            cin >> u >> v >> w;
             if (g[u][v] > w)
                g[u][v] = g[v][u] = w;
        }
         //   Input
        ans = INF;
        memset(path, 0 , sizeof (path));
         //   Init
         for ( long  i = 1 ;i <= n;i ++ )
             for ( long  j = i + 1 ;j <= n;j ++ )
            {
                 long  t = g[i][j],now;
                g[i][j] = g[j][i] = INF;
                now = dijkstra(i,j) + t;
                 if (ans > now)
                {
                    ans = now;
                     //   Update
                    num = 0 ;
                    memset(path, 0 , sizeof (path));
                     long  r = j;
                     while (r)
                    {
                        num ++ ;
                        path[num] = r;
                        r = tmp[r];
                    }
                     //   Deal path
                }
                g[i][j] = g[j][i] = t;
            }
         //   Find the shorest path
         if (ans >= INF)
            cout << " No solution. " << endl;
         else
        {
             for ( long  i = num;i >= 1 ;i -- )
            {
                 if (i != num) cout << "   " ;
                cout << path[i];
            }
            cout << endl;
        }
    }
return   0 ;
}