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\n", 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\n", *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