1021 Deepest Root （25 分）

最新推荐文章于 2023-07-28 14:02:44 发布

原创最新推荐文章于 2023-07-28 14:02:44 发布 · 265 阅读

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本文探讨了在给定的树形结构中，如何找到使得树高度最大的根节点，即最深根。通过深度优先搜索算法，文章详细解释了如何遍历树的每一个节点，比较并记录下高度最大的节点。此外，还讨论了当输入的图不是一个树的情况下的错误处理策略。

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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 (≤104) 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:

Sample Output 1:

3
4
5

Sample Input 2:

Sample Output 2:

Error: 2 components

#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<set>
#include<map>
#include<vector>
#include<string>
using namespace std;
typedef long long ll;
map<int ,vector<int> >mymap;
vector<int> v;
set<int> s;
int mm;
int vis[100002];
void dfs(int x,int step)
{
    vis[x] = 1;
    if(step>mm){
        mm=step;
        v.clear();
        v.push_back(x);
    }
    else if(step==mm){
        v.push_back(x);
    }
    for(int i = 0 ; i<mymap[x].size();i++)
    {
        if(!vis[mymap[x][i]]){
            dfs(mymap[x][i],step+1);
        }
    }
}
int main(){
    int N;
    cin>>N;
    for(int i = 1; i <=N-1;i++)
    {
        int a,b;
        cin>>a>>b;
        mymap[a].push_back(b);
        mymap[b].push_back(a);
    }
    int temp = 0;
    int k = 0;
    for(int i = 1; i<=N;i++)
    {
        if(!vis[i]){
            dfs(i,0);
            if(i==1){
                for(int j = 0;j<v.size();j++)
                {
                    if(j==0)temp = v[j];
                    s.insert(v[j]);
                }
            }
            k++;//计算连通分量
        }
    }
    if(k==1){
        memset(vis,0,sizeof(vis));
        v.clear();
        dfs(temp,0);
        for(int i = 0;i<v.size();i++)
        {
            s.insert(v[i]);
        }
        set<int>::iterator it;
        for(it = s.begin();it!=s.end();it++)
        {
            printf("%d\n",*it);
        }
    }
    else printf("Error: %d components\n",k );
    return 0;
}