PAT甲级1021

1021 Deepest Root （25 分）

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 (≤10​4​​) 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

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

const int Max = 10010;

int n, k = 0, maxh = -1;
vector<int> node[Max];
bool is[Max] = {false};
vector<int> temp;
set<int> ans;

void dfs(int x){
	is[x] = true;
	for(int i = 0; i < node[x].size(); i++){
		if(!is[node[x][i]])
			dfs(node[x][i]);
	}
}

void conn(){
	for(int i = 1; i <= n; i++){
		if(!is[i]){
			dfs(i);
			k++;
		}
	}
}

void deep(int x, int high, int pre){
	if(high > maxh){
		temp.clear();
		maxh = high;
		temp.push_back(x);
	}
	else if(high == maxh){
		temp.push_back(x);
	}
	for(int i = 0; i < node[x].size(); i++){
		if(node[x][i] == pre)
			continue;
		deep(node[x][i], high + 1, x);
	}
}

int main(){
	scanf("%d", &n);
	int a, b;
	for(int i = 0; i < n - 1; i++){
		scanf("%d %d", &a, &b);
		node[a].push_back(b);
		node[b].push_back(a);
	}
	conn();
	if(k > 1){
		printf("Error: %d components", k);
		return 0;
	}
	deep(1, 0, 0);
	for(int i = 0; i < temp.size(); i++)
		ans.insert(temp[i]);
	maxh = -1;
	a = temp[0];
	temp.clear();
	deep(a, 0, 0);
	for(int i = 0; i < temp.size(); i++)
		ans.insert(temp[i]);
	for(set<int>::iterator it = ans.begin(); it != ans.end(); it++)
		printf("%d\n", *it);
	return 0;
}