[hiho 11]树的直径

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题目描述

树的直径必然取自某个节点的两个（或一个）最深的子树，可以一遍DFS解决。

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

vector<int> tree[100005];
bool has_parent[100005] = { false };
int max_length[100005];
int ans;

void solve(int root_idx) {
	int lgst = 0, lgst_idx = 0;
	int size = tree[root_idx].size();
	for (int i = 0; i < size; i++) { // 递归求树深
		int sub_idx = tree[root_idx][i];
		solve(sub_idx);
		if (lgst < max_length[sub_idx]) {
			lgst = max_length[sub_idx];
			lgst_idx = i;
		}
	}
	max_length[root_idx] = lgst + 1;

	int sec_lgst = 0;
	for (int i = 0; i < size; i++) { // 次深子树深度
		int sub_idx = tree[root_idx][i];
		if (sec_lgst < max_length[sub_idx] && i != lgst_idx) {
			sec_lgst = max_length[sub_idx];
		}
	}
	ans = max(ans, lgst + sec_lgst);
}

int main() {
	int n;
	cin >> n;
	for(int i = 1; i < n; i++) {
		int a, b;
		cin >> a >> b;
		if (has_parent[a]) {
			tree[a].push_back(b);
			has_parent[b] = true;
		} else {
			tree[b].push_back(a);
			has_parent[a] = true;
		}
	}
	for (int i = 1; i <= n; i++) {
		if (!has_parent[i]) {
			solve(i);
			break;
		}
	}

	cout << ans << endl;
	return 0;
}

另一种做法是两遍BFS：以任一个节点为根BFS找到最远的节点，再从该节点BFS找到的最长路径就是直径。

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>

using namespace std;

vector<int> tree[100005];
int length[100005];
int max_length, max_idx;

void bfs(int root_idx, int pa_idx, int depth) {
	length[root_idx] = depth;
	if (max_length < depth) {
		max_length = depth;
		max_idx = root_idx;
	}
	int len = tree[root_idx].size();
	for (int i = 0; i < len; i++) {
		int sub_idx = tree[root_idx][i];
		if (sub_idx != pa_idx) {
			bfs(sub_idx, root_idx, depth + 1);
		}
		
	}
}

int main() {
	int n;
	cin >> n;
	for(int i = 1; i < n; i++) {
		int a, b;
		cin >> a >> b;
		tree[a].push_back(b);
		tree[b].push_back(a);
	}
	bfs(1, -1, 0);
	bfs(max_idx, -1, 0);
	cout << max_length << endl;
	return 0;
}

[hiho 11]树的直径

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原文地址：http://www.cnblogs.com/xblade/p/4478218.html