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A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N-1 lines follow, each describes an edge by given the two adjacent nodes‘ numbers.
Output Specification:
For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print "Error: K components" where K is the number of connected components in the graph.
Sample Input 1:
5 1 2 1 3 1 4 2 5
Sample Output 1:
3 4 5
Sample Input 2:
5 1 3 1 4 2 5 3 4
Sample Output 2:
Error: 2 components
#include <stdio.h> #include <stdlib.h> #include <iostream> #include <string.h> #include <math.h> #include <algorithm> #include <string> #include <stack> #include <queue> #include <vector> using namespace std; const int maxn=100010; vector<int> G[maxn]; int father[maxn]; //并查集初始化 void init() { for(int i=1;i<maxn;i++) { father[i]=i; } } int findFather(int x) { int a=x; while(x!=father[x]) { x=father[x]; } //路径压缩 while(a!=father[a]) { int tmp=a; a=father[a]; father[tmp]=x; } return x; } void Union(int a,int b)//合并 { int faA=findFather(a); int faB=findFather(b); if(faA!=faB) { father[faA]=faB; } } //计算集合数目 int isRoot[maxn]; int cal(int n) { int block=0; for(int i=1;i<=n;i++) { isRoot[findFather(i)]=true; } for(int i=1;i<=n;i++) { block+=isRoot[i]; } return block; } //以某节点为根，遍历 int maxH=0; vector<int> tmp,Ans; void DFS(int u,int height,int pre) { if(height>maxH) { tmp.clear(); tmp.push_back(u); maxH=height; } else if(height==maxH) { tmp.push_back(u); } for(int i=0;i<G[u].size();i++) { if(G[u][i]==pre)continue; DFS(G[u][i],height+1,u); } } int main(){ int a,b,n; scanf("%d",&n); init(); for(int i=1;i<n;i++)//从一开始 { scanf("%d%d",&a,&b);//无向图，节点用vector存储 G[a].push_back(b); G[b].push_back(a); Union(a,b); } int block=cal(n); if(block!=1) { printf("Error: %d components",block); }else { DFS(1,1,-1); Ans=tmp; DFS(Ans[0],1,-1); for(int i=0;i<tmp.size();i++) { Ans.push_back(tmp[i]); } sort(Ans.begin(),Ans.end()); printf("%d\n",Ans[0]); for(int i=1;i<Ans.size();i++) { if(Ans[i]!=Ans[i-1]) { printf("%d\n",Ans[i]); } } } return 0; }
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原文地址：http://www.cnblogs.com/ligen/p/4323685.html
