1021. Deepest Root (25)
A graph which is connected and acyclic can be considered a tree. The height 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 (<=10000) 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 5Sample Output 1:
3 4 5Sample Input 2:
5 1 3 1 4 2 5 3 4Sample Output 2:
Error: 2 components
思路
二次dfs
1.第一次dfs确定图是否连通,如果连通找到最深的那个点first(多个就取最先被找到的),没有输出题目要求的错误信息。
2.第二次dfs重置所有状态,然后从first开始dfs,找到的所有的最深的点即是题目要求的节点,依次插入一个set容器中(每次插入会自动排序)。
3.输出set中的所有元素就行。
代码
#include<iostream> #include<vector> #include<set> using namespace std; vector<vector<int>> graph; vector<int> highestNodes; vector<bool> visits(10005,false); int maxheight = 1; set<int> results; void dfs(int root,int height) { visits[root] = true; if(height >= maxheight) { if(height > maxheight) { highestNodes.clear(); maxheight = height; } highestNodes.push_back(root); } for(int i = 0;i < graph[root].size();i++) { if(!visits[graph[root][i]]) dfs(graph[root][i],height + 1); } } inline void resetVisits(const int n) { for(int i = 0;i <= n;i++) visits[i] = false; } int main() { int N; while(cin >> N) { //input graph.resize(N + 1); for(int i = 1;i < N;i++) { int a,b; cin >> a >> b; graph[b].push_back(a); graph[a].push_back(b); } int cnt = 0; //handle int first = -1; //最高的节点之一 for(int i = 1;i <= N;i++) { if(!visits[i]) { cnt++; dfs(i,1); if( i == 1) { for(int j = 0;j < highestNodes.size();j++) { results.insert(highestNodes[j]); if(j == 0) first = highestNodes[j]; } } } } if(cnt > 1) cout << "Error: "<< cnt <<" components" << endl; else { highestNodes.clear(); maxheight = 1; resetVisits(N); dfs(first,1); for(int i = 0;i < highestNodes.size();i++ ) results.insert(highestNodes[i]); for(auto it = results.begin();it != results.end();it++) cout << *it << endl; } } }