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请找出无向图中相连要素的个数。
图中的每个节点包含其邻居的 1 个标签和 1 个列表。(一个无向图的相连节点(或节点)是一个子图,其中任意两个顶点通过路径相连,且不与超级图中的其它顶点相连。)
样例
给定图:
A------B C
\ | |
\ | |
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D E
返回 {A,B,D}, {C,E}
。其中有 2 个相连的元素,即{A,B,D}, {C,E}
思路:说实话,lintcode的翻译我并没有看懂是啥意思,当然更不知道怎么做。准确的说,这个问题应该叫做在无向图中找各极大连通子图,使用的算法是广度优先搜索。此处附上广度优先搜索和深度优先搜索算法的比较,顺带复习一下;
http://blog.csdn.net/andyelvis/article/details/1728378
当然代码也是我学习他人的,来源附上http://www.jiuzhang.com/solutions/find-the-connected-component-in-the-undirected-graph/
1 /** 2 * Definition for Undirected graph. 3 * class UndirectedGraphNode { 4 * int label; 5 * ArrayList<UndirectedGraphNode> neighbors; 6 * UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); } 7 * }; 8 */ 9 public class Solution { 10 /** 11 * @param nodes a array of Undirected graph node 12 * @return a connected set of a Undirected graph 13 */ 14 public List<List<Integer>> connectedSet(ArrayList<UndirectedGraphNode> nodes) { 15 // Write your code here 16 int m = nodes.size(); 17 Map<UndirectedGraphNode,Boolean> visited = new HashMap<>(); 18 for(UndirectedGraphNode node:nodes){ 19 visited.put(node,false); 20 } 21 List<List<Integer>> result = new ArrayList<>(); 22 for(UndirectedGraphNode node:nodes){ 23 if(visited.get(node) == false){ 24 bfs(node,visited,result); 25 } 26 } 27 return result; 28 } 29 public void bfs(UndirectedGraphNode node,Map<UndirectedGraphNode,Boolean> visited,List<List<Integer>> result){ 30 List<Integer> row = new ArrayList<>(); 31 Queue<UndirectedGraphNode> queue = new LinkedList<>(); 32 visited.put(node,true); 33 queue.offer(node); 34 while(!queue.isEmpty()){ 35 UndirectedGraphNode u = queue.poll(); 36 row.add(u.label); 37 for(UndirectedGraphNode v:u.neighbors){ 38 if(visited.get(v)==false){ 39 visited.put(v,true); 40 queue.offer(v); 41 } 42 } 43 } 44 Collections.sort(row); 45 result.add(row); 46 } 47 }
找出无向图汇总的相连要素find-the-connected-component-in-the-undirected-graph
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原文地址:http://www.cnblogs.com/wangnanabuaa/p/5014835.html