标签：
https://leetcode.com/problems/minimum-height-trees/
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;
}
};
标签：
原文地址：http://www.cnblogs.com/GadyPu/p/5011109.html
