1021 Deepest Root （25 分）

A graph which is connected and acyclic can be considered a tree. The hight 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 (≤10​⁴​​ ) 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

分析

• 首先从任意节点通过dfs得到连通分量的个数

for(node in node_set){
   if(visit[node]==false) dfs(node);
   cnt++;
}

• 在dfs函数中设置参数保存当前长度，递归时深度加1，求取当前最大深度，并将这些节点保存到temp向量中

void dfs(int node,int height){
    if(height>maxheight){
        temp.clear();
        temp.push_back(node);
        maxheight=height;
    }else if(height==maxheight){
        temp.push_back(node);
    }
    ...
}

• dfs一次后把temp中元素插入集合s
• 在这些节点中任意一个进行dfs，得到的节点跟上一步的节点的并集即为结果。

#include 
#include 
#include 
#include 
using namespace std;

int n;
int maxheight=0;


vector v[10010];
vector temp;
bool visit[10010];
set s;
void dfs(int node,int height){
    if(height>maxheight){
        temp.clear();
        temp.push_back(node);
        maxheight=height;
    }else if(height==maxheight){
        temp.push_back(node);
    }
    visit[node]=true;
    for(int i=0;i

参考