zoj3204 Connect them 最小生成树

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3367

题目就是简单的最小生成树的模板的应用，不过最小生成树可能不唯一，答案要求输出字典序最小

代码：

 #include<cstdlib>
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define maxn 110
 int n,tol;
 int cnt;
 class node
 {
     public:
     int from;
     int to;
     int w;
 };
 node edge[maxn*maxn];
 node ans[maxn*maxn];
 int parent[maxn*maxn];
 void addedge(int u,int v,int w)
 {
     edge[tol].from=u;
     edge[tol].to=v;
     edge[tol].w=w;
     tol++;
 }
 void UFset()
 {
     for(int i=;i<maxn*maxn;i++)
        parent[i]=-;
 }
 int Find( int x)
 {
    int s;
    for(s=x;parent[s]>=; s=parent[s]);
    while(s!=x)
    {
       int tmp=parent[x];
       parent[x]=s;
       x=tmp;
    }
    return s;
 }
 void Union(int R1, int R2)
 {
     int root1=Find(R1);
     int root2=Find(R2);

     int tmp=parent[root1]+parent[root2];

     if(parent[root1]> parent[root2])
      {
          parent[root1]=root2;
          parent[root2]=tmp;
      }
      else
      {
          parent[root2]=root1;
          parent[root1]=tmp;
      }
 }
 bool cmp1( node a, node b)
 {
     if(a.w!=b.w)return a.w<b.w;
        else if(a.from!=b.from)return a.from<b.from;
           else return a.to<b.to;
 }
 bool cmp2(node a, node b)
 {
   if(a.from!=b.from)return a.from<b.from;
      else return a.to<b.to;
 }
 void Kruskal()
 {
    int num=;
    int u,v;
    UFset();
    cnt=;

    for(int i=;i<tol;i++)
     {
        u=edge[i].from;
        v=edge[i].to;
        if(Find(u) != Find(v))
        {
            ans[cnt++]=edge[i];
            Union(u,v);
        }

     }
 }
 int main()
 {
      int t;
      scanf("%d",&t);
      while(t--)
      {
         scanf("%d",&n);
         tol=;
         int  weight;
         for(int i=;i<=n;i++)
           for(int j=;j<=n;j++)
            {
                 scanf("%d",&weight);
                 if(j<=i) continue;
                 if(weight== ) continue;
                 addedge(i,j,weight);
            }
          sort(edge,edge+tol,cmp1);

          Kruskal();

          if(cnt!=n-)
               cout<<"-1"<<endl;
            else
               {
                   sort(ans,ans+cnt,cmp2);
                   cout<<ans[].from<<" "<<ans[].to;
                   for(int i=;i<cnt;i++)
                       cout<<" "<<ans[i].from<<" "<<ans[i].to;
                       cout<<endl;
               }
      }
      return ;
 }