1021. Deepest Root

最新推荐文章于 2024-10-11 16:39:47 发布

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最新推荐文章于 2024-10-11 16:39:47 发布

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本文深入探讨了深度学习和数据挖掘领域的关键概念、技术应用与最新进展，包括机器学习模型、算法优化、数据预处理、特征工程、模型评估等方面的内容。通过实例分析，展示了深度学习在图像识别、自然语言处理、推荐系统等领域的强大能力，同时讨论了数据挖掘在市场分析、客户行为预测、异常检测等场景中的实际应用。

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1021. Deepest Root (25)

时间限制

1500 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

Sample Output 1:

3
4
5

Sample Input 2:

Sample Output 2:

Error: 2 components

考查：并查集、树的深度。

#include<stdio.h>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;

vector<int> v[10002];
queue<int> que;
int Tree[10002];
int ans[10002];
bool mark[10002];
int tail, n;

int find_root(int x)
{
	if(Tree[x] == -1)
		return x;
	else
	{
		int tmp = find_root(Tree[x]);
		Tree[x] = tmp;
		return tmp;
	}
}

int find_height(int x)
{
	mark[x] = true;
	int height_tmp = 0;
	que.push(x);
	que.push(-1);
	
	while(!que.empty())
	{
		if(que.front() == -1)
		{
			height_tmp ++;
			que.pop();
			
			if(que.empty())
				break;
			else
				que.push(-1);
		}
		int front = que.front();
		tail = front;
		
		for(int i = 0; i < v[front].size(); i ++)
		{
			if(!mark[v[front][i]])
				que.push(v[front][i]);
			mark[v[front][i]] = true;
		}
		que.pop();
	}
	for(int i = 1; i <= n; i ++)
		mark[i] = false;
	return height_tmp;
}

int main()
{
	freopen("F://Temp/input.txt", "r", stdin);
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++)
	{
		Tree[i] = -1;
		ans[i] = 0;
		mark[i] = false;
	}
	
	if(!que.empty())
		que.pop();
		
	int a, b;
	for(int i = 0; i < n-1; i ++)
	{
		scanf("%d%d", &a, &b);
		v[a].push_back(b);
		v[b].push_back(a);
		a = find_root(a);
		b = find_root(b);
		if(a != b)
			Tree[a] = b;
	}
	int count = 0;
	for(int i = 1; i <= n; i ++)
		if(Tree[i] == -1)
			count ++;
	if(count > 1)
	{
		printf("Error: %d components\n", count);
		return 0;
	}
	else
	{
		int height_max, head;
		for(int i = 1; i <= n; i ++)
		{
			if(v[i].size() == 1)
			{
				head = i;
				break;
			}
		}
		
		height_max = find_height(head);
		while(height_max != find_height(tail))
		{
			head = tail;
			height_max = find_height(head);
		}
		int j = 0;
		for(int i = 1; i <= n; i ++)
			if(find_height(i) == height_max)
				ans[j ++] = i;
	//	printf("height_max = %d\n", height_max);
	//	printf("j = %d\n", j);
		sort(ans, ans + j);
		for(int i = 0; i < j; i ++)
			printf("%d\n", ans[i]);
	}
	return 0;
}