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 N (≤) 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

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