【PAT】A1021. Deepest Root (25)

A1021. 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 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

 //NKW 甲级练习题1032
#pragma warning(disable:4996)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
#include <algorithm>
#define INF 1000000000
using namespace std;
const int maxn = 10010;
int n, u, v, depthest[maxn], k = 0, dep = 0;
bool vis[maxn];
vector<int > adj[maxn], roots;
void DFS(int u, int depth, int root){
	vis[u] = 1;
	if (depth > depthest[root])
		depthest[root] = depth;
	for (int i = 0; i < adj[u].size(); i++)
		if (vis[adj[u][i]] == false)
			DFS(adj[u][i], depth + 1, root);
}
void DFSTravel(){
	for (int i = 1; i <= n; i++){
		if (vis[i] == false){
			DFS(i, 0, i);
			k++;
		}
	}
}
void DFSTravel2(){
	for (int i = 2; i <= n; i++){
		memset(vis, 0, sizeof(vis));
		if (vis[i] == false)
			DFS(i, 0, i);
	}
}
int main(){
	scanf("%d", &n);
	for (int i = 1; i < n; i++){
		scanf("%d %d", &u, &v);
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	DFSTravel();
	if (k>1)
		printf("Error: %d components\n", k);
	else{
		DFSTravel2();
		for (int i = 1; i <= n; i++)
			if (depthest[i] > dep)
				dep = depthest[i];
		for (int i = 1; i <= n; i++)
			if (dep == depthest[i])
				roots.push_back(i);
		sort(roots.begin(), roots.end());
		for (int i = 0; i < roots.size(); i++)
			printf("%d\n", roots[i]);
	}

	system("pause");
	return 0;
}