[LintCode] Connecting Graph II

最新推荐文章于 2022-11-10 15:03:08 发布

转载最新推荐文章于 2022-11-10 15:03:08 发布 · 123 阅读

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原文链接：http://www.cnblogs.com/lz87/p/7500090.html

本文介绍了一种使用并查集数据结构处理图论中节点连接查询的问题。具体实现了一个UnionFind类来维护节点之间的连接关系，并通过connect方法添加边，query方法返回指定节点所在连通分量的节点数。

Given n nodes in a graph labeled from 1 to n. There is no edges in the graph at beginning.

You need to support the following method:
1. connect(a, b), an edge to connect node a and node b
2. query(a), Returns the number of connected component nodes which include node a.

Example

5 // n = 5
query(1) return 1
connect(1, 2)
query(1) return 2
connect(2, 4)
query(1) return 3
connect(1, 4)
query(1) return 3

 1 class UnionFind {
 2     private int[] father = null;
 3     private int[] numsInUnion = null;
 4     
 5     public UnionFind(int n){
 6         father = new int[n];
 7         numsInUnion = new int[n];
 8         for(int i = 0; i < n; i++){
 9             father[i] = i;
10             numsInUnion[i] = 1;
11         }
12     }
13     public int find(int x){
14         if(father[x] == x){
15             return x;
16         }
17         return father[x] = find(father[x]);
18     }
19     public void connect(int a, int b){
20         int root_a = find(a);
21         int root_b = find(b);
22         if(root_a != root_b){
23             father[root_a] = root_b;
24             numsInUnion[root_b] += numsInUnion[root_a];
25         }
26     }
27     public int queryNumInUnion(int a){
28         return numsInUnion[find(a)];
29     }
30 }
31 public class ConnectingGraph2 {
32     private UnionFind uf = null;
33     public ConnectingGraph2(int n) {
34         uf = new UnionFind(n);
35     }
36 
37     public void connect(int a, int b) {
38         uf.connect(a - 1, b - 1);
39     }
40         
41     public int query(int a) {
42         return uf.queryNumInUnion(a - 1);
43     }
44 }