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
希望新年心里多一点温暖吧 失去时间就失去 再恨再遗憾也是不会回来 向前看吧 记新年熬第一夜
FH
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