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

生词

题目大意：

给出n个结点（1~n）之间的n条边，问是否能构成一棵树，如果不能构成则输出它有的连通分量个数，如果能构成一棵树，输出能构成最深的树的高度时，树的根结点。如果有多个，按照从小到大输出。

分析：

首先深度优先搜索判断它有几个连通分量。如果有多个，那就输出Error: x components，如果只有一个，就两次深度优先搜索，先从一个结点dfs后保留最高高度拥有的结点们，然后从这些结点中的其中任意一个开始dfs得到最高高度的结点们，这两个结点集合的并集就是所求

原文链接：https://blog.csdn.net/liuchuo/article/details/52294178

题解

咱就是说，这段证明谁能看懂呢。。。（除了各位大佬





1.用set确实更方便：自动去重+自动排序！
2.并查集模板容易忘写init函数。

#include <bits/stdc++.h>
using namespace std;
const int N=100010;
vector<int> G[N];

bool isRoot[N];
int father[N];
int findFather(int x)
{
    int a=x;
    while(x!=father[x]){
        x=father[x];
    }
    while(a!=father[a]){
        int z=a;
        a=father[a];
        father[z]=x;
    }
    return x;
}
void Union(int a,int b)
{
    int faA=findFather(a);
    int faB=findFather(b);
    if(faA!=faB){
        father[faA]=faB;
    }
}
void init(int n)
{
    for(int i=1;i<=n;i++){
        father[i]=i;
    }
}
int calBlock(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;
}
int maxH=0;
set<int> temp,Ans;
void DFS(int u,int Height,int pre)
{
    if(Height>maxH){
        temp.clear();
        temp.insert(u);
        maxH=Height;
    }else if(Height==maxH){
        temp.insert(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;
    cin>>n;
    init(n);
    for(int i=1;i<n;i++){
        cin>>a>>b;
        G[a].push_back(b);
        G[b].push_back(a);
        Union(a,b);
    }
    int Block=calBlock(n);
    if(Block!=1)
        cout<<"Error: "<<Block<<" components"<<endl;
    else{
        DFS(1,1,-1);
        Ans=temp;
        auto it=Ans.begin();
        DFS(*it,1,-1);
        for(auto it=temp.begin();it!=temp.end();it++){
            Ans.insert(*it);
        }
        for(auto it=Ans.begin();it!=Ans.end();it++)
            cout<<*it<<endl;
    }
	return 0;
}

标签：case,25,结点,1021,temp,int,father,dfs,root
来源： https://www.cnblogs.com/moonlight1999/p/15877247.html