1021 Deepest Root （25 分）

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.

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思路如下：

1. 判断是否为树。由于题目已经表明，只有N-1条边，所以必然是数，或者是带环路的的森林，这样问题就转换为判断连通块的个数。
2. 第二个问题就是计算最大深度的问题，由于题目要求输出所有的最大深度的点，这里使用distance[k]来表示k点为root的最大深度。再设置一个index[k]数组，判断k时候为最大值。最后遍历index数组，输出所有的最大值的序号。

下面是AC的 代码：
这里没有使用using namespace std; 是因为会和全局变量冲突

#include <stdio.h>
#include <algorithm>
#include <vector>
using std::vector;
using std::fill;
const int maxn = 1e4 + 10;
vector<vector<int>> G;
bool isvisit[maxn];
int n, distance[maxn] = {0};

void DFS(int index) {
	isvisit[index] = true;
	if (G[index].size() == 0) return;
	else {
		for (int i = 0; i < G[index].size(); i++) {
			if (isvisit[G[index][i]] == false)
				DFS(G[index][i]);
		}
	}
}
//计算index为root节点的最远距离
void cal(int index,int height,int k) {  //k为当前节点
	isvisit[k] = true;
	height++;
	if (height > distance[index])
		distance[index] = height;
	for (int i = 0; i < G[k].size(); i++) {
		if(isvisit[G[k][i]]==false)
			cal(index, height, G[k][i]);
	}
	return;
}

int main() {
	fill(isvisit, isvisit + maxn, false);
	int a,b;
	scanf("%d", &n);
	G.resize(n + 1);
	for (int i = 1; i < n; i++) {
		scanf("%d %d", &a, &b);
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int cnt = 0;
	for (int i = 1; i <= n; i++) {
		if (isvisit[i] == false) {
			cnt++;
			DFS(i);
		}
	} //连通块
	if (cnt > 1) {
		printf("Error: %d components", cnt);
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		fill(isvisit, isvisit + maxn, false);
		cal(i, 0, i);
	}
	
	vector<bool> ismax(n + 1);
	fill(ismax.begin(), ismax.end(), true);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (distance[i] < distance[j]) {
				ismax[i] = false;
				break;
			}
		}
	}
	for (int i = 1; i <= n; i++)
		if (ismax[i]) printf("%d\n", i);
	return 0;
}