PAT 甲级 1021 Deepest Root

https://pintia.cn/problem-sets/994805342720868352/problems/994805482919673856

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

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

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N;
vector<int> v[maxn];
int vis[maxn], mp[maxn];
int cnt = 0;
int depth = INT_MIN;
vector<int> ans;

void dfs(int st) {
    vis[st] = 1;

    for(int i = 0; i < v[st].size(); i ++) {
        if(vis[v[st][i]] == 0)
            dfs(v[st][i]);
    }
}

void helper(int st, int step) {
    if(step > depth) {
        ans.clear();
        ans.push_back(st);
        depth = step;
    } else if(step == depth) ans.push_back(st);

    mp[st] = 1;
    for(int i = 0; i < v[st].size(); i ++) {
        if(mp[v[st][i]] == 0)
            helper(v[st][i], step + 1);
    }
}

int main() {
    scanf("%d", &N);
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < N - 1; i ++) {
        int a, b;
        scanf("%d%d", &a, &b);
        v[a].push_back(b);
        v[b].push_back(a);
    }

    for(int i = 1; i <= N; i ++) {
        if(vis[i] == 0) {
            dfs(i);
            cnt ++;
        }
        else continue;
    }

    set<int> s;
    int beginn = 0;
    helper(1, 1);
    if(ans.size() != 0) beginn = ans[0];
    for(int i = 0; i < ans.size(); i ++)
        s.insert(ans[i]);

    if(cnt >= 2)
        printf("Error: %d components
", cnt);
    else {
        ans.clear();
        depth = INT_MIN;
        memset(mp, 0, sizeof(mp));
        helper(beginn, 1);
        for(int i = 0; i < ans.size(); i ++)
            s.insert(ans[i]);

        for(set<int>::iterator it = s.begin(); it != s.end(); it ++)
            printf("%d
", *it);
    }
    return 0;
}

　　第一个 dfs 搜索有多少个连通块 helper 来找树的直径的一个头已知树的直径树上任意一点到的最大距离的另一端一定是树的直径的一个端点两次深搜