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：
1 #include<cstdlib> 2 #include<cstdio> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 using namespace std; 7 #define maxn 110 8 int n,tol; 9 int cnt; 10 class node 11 { 12 public: 13 int from; 14 int to; 15 int w; 16 }; 17 node edge[maxn*maxn]; 18 node ans[maxn*maxn]; 19 int parent[maxn*maxn]; 20 void addedge(int u,int v,int w) 21 { 22 edge[tol].from=u; 23 edge[tol].to=v; 24 edge[tol].w=w; 25 tol++; 26 } 27 void UFset() 28 { 29 for(int i=0;i<maxn*maxn;i++) 30 parent[i]=-1; 31 } 32 int Find( int x) 33 { 34 int s; 35 for(s=x;parent[s]>=0; s=parent[s]); 36 while(s!=x) 37 { 38 int tmp=parent[x]; 39 parent[x]=s; 40 x=tmp; 41 } 42 return s; 43 } 44 void Union(int R1, int R2) 45 { 46 int root1=Find(R1); 47 int root2=Find(R2); 48 49 int tmp=parent[root1]+parent[root2]; 50 51 if(parent[root1]> parent[root2]) 52 { 53 parent[root1]=root2; 54 parent[root2]=tmp; 55 } 56 else 57 { 58 parent[root2]=root1; 59 parent[root1]=tmp; 60 } 61 } 62 bool cmp1( node a, node b) 63 { 64 if(a.w!=b.w)return a.w<b.w; 65 else if(a.from!=b.from)return a.from<b.from; 66 else return a.to<b.to; 67 } 68 bool cmp2(node a, node b) 69 { 70 if(a.from!=b.from)return a.from<b.from; 71 else return a.to<b.to; 72 } 73 void Kruskal() 74 { 75 int num=0; 76 int u,v; 77 UFset(); 78 cnt=0; 79 80 for(int i=0;i<tol;i++) 81 { 82 u=edge[i].from; 83 v=edge[i].to; 84 if(Find(u) != Find(v)) 85 { 86 ans[cnt++]=edge[i]; 87 Union(u,v); 88 } 89 90 } 91 } 92 int main() 93 { 94 int t; 95 scanf("%d",&t); 96 while(t--) 97 { 98 scanf("%d",&n); 99 tol=0; 100 int weight; 101 for(int i=1;i<=n;i++) 102 for(int j=1;j<=n;j++) 103 { 104 scanf("%d",&weight); 105 if(j<=i) continue; 106 if(weight==0 ) continue; 107 addedge(i,j,weight); 108 } 109 sort(edge,edge+tol,cmp1); 110 111 Kruskal(); 112 113 if(cnt!=n-1) 114 cout<<"-1"<<endl; 115 else 116 { 117 sort(ans,ans+cnt,cmp2); 118 cout<<ans[0].from<<" "<<ans[0].to; 119 for(int i=1;i<cnt;i++) 120 cout<<" "<<ans[i].from<<" "<<ans[i].to; 121 cout<<endl; 122 } 123 } 124 return 0; 125 }