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最新推荐文章于 2024-10-11 16:39:47 发布

原创最新推荐文章于 2024-10-11 16:39:47 发布 · 784 阅读

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1021. Deepest Root (25)

时间限制

1500 ms

内存限制

32000 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

用了个很土的办法啊，最后一个测试案例内存超限了。

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

int map[10001][10001], n;
int deep[10010], vis[10010], maxx;

void dfs(int t, int ll)
{
	int i;

	vis[t] = 1;

	if(ll > maxx)
		maxx = ll;

	for(i = 1; i <= n; i++)
	{
		if(vis[i] == 0 && map[t][i] == 1)
			dfs(i, ll+1);
	}
}

int main()
{
	int i, j, a, b;

	while(scanf("%d", &n) != EOF)
	{
		for(i = 1; i < n; i++)
		{
			scanf("%d %d", &a, &b);
			map[a][b] = 1;
			map[b][a] = 1;
		}

		memset(deep, 0, sizeof(deep));

		int tmp = 0;

		for(i = 1; i <= n; i++)
		{
			memset(vis, 0, sizeof(vis));
			maxx = -1;
			dfs(i, 1);
			deep[i] = maxx;

			for(j = 1; j <= n; j++)
				if(vis[j] == 0)
					tmp++;
		}

		if(tmp != 0)
		{
			int count = 1;

			for(i = 1; i <= n; i++)
			{
				if(vis[i] == 0)
				{
					dfs(i, 0);
					count++;
				}
			}

			printf("Error: %d components\n", count);

			continue;
		}

		for(i = 1; i <= n; i++)
			if(deep[i] > tmp)
				tmp = deep[i];

		for(i = 1; i <= n; i++)
		{
			if(deep[i] == tmp)
				printf("%d\n", i);
		}
	}
	return 0;
}