leetcode Minimum Height Trees

题目连接

https://leetcode.com/problems/minimum-height-trees/  

Minimum Height Trees

Description

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / 
      2   3

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
       | /
        3
        |
        4
        |
        5

return [3, 4]

题目大意：给你一张无向图，可以选择任意节点作为根使其成为有根树。找到一些节点 使得这棵树的高度最小。。
我的思路：先找到树中的最长链，其中点(节点)即为所求。。
找树中的最长链两次bfs即可。。

class Solution {
private:
	typedef vector<int> vec;
	typedef vector<pair<int, int>> vpi;
public:
	vec findMinHeightTrees(int n, vpi& edges) {
		tot = 0, ret.clear();
		if (edges.empty()) { ret.push_back(0); return ret; }
		init(n, edges);
		vec ans = solve(n);
		__free__();
		return ans;
	}
private:
	vec ret;
	int tot, *head, *dist, *pre;
	struct edge { int to, next; }*G;
	inline void init(int n, vpi& edges) {
		int m = n + 10;
		pre = new int[m];
		head = new int[m];
		dist = new int[m];
		memset(pre, -1, sizeof(int)* m);
		memset(head, -1, sizeof(int)* m);
		m = edges.size();
		G = new edge[(m + 10) << 1];
		for (int i = 0; i < m; i++) {
			int u = edges[i].first, v = edges[i].second;
			add_edge(u, v);
		}
	}
	inline void add_edge(int u, int v) {
		G[tot].to = v, G[tot].next = head[u], head[u] = tot++;
		G[tot].to = u, G[tot].next = head[v], head[v] = tot++;
	}
	inline int bfs(int s, int n, bool f = false) {
		int id = s, max_dist = 0;
		memset(dist, -1, sizeof(int) * (n + 10));
		queue<int> q; q.push(s);
		dist[s] = 0;
		while (!q.empty()) {
			int u = q.front(); q.pop();
			if (dist[u] > max_dist) {
				max_dist = dist[id = u];
			}
			for (int i = head[u]; ~i; i = G[i].next) {
				int &v = G[i].to;
				if (-1 == dist[v]) {
					dist[v] = dist[u] + 1;
					if (f) pre[v] = u;
					q.push(v);
				}
			}
		}
		return id;
	}
	inline vec solve(int n) {
		int s = bfs(0, n);
		int t = bfs(s, n, true);
		vec ans;
		for (; ~t; t = pre[t]) ret.push_back(t);
		n = ret.size();
		if (!n) return ans;
		ans.push_back(ret[n / 2]);
		if (!(n & 1)) ans.push_back(ret[n / 2 - 1]);
		if (ans.size() > 1 && ans[0] > ans[1]) swap(ans[0], ans[1]);
		return ans;
	}
	inline void __free__() {
		delete []G; delete []pre;
		delete []dist; delete []head;
	}
};