zoukankan      html  css  js  c++  java
  • Lintcode: Topological Sorting

    Given an directed graph, a topological order of the graph nodes is defined as follow:
    
    For each directed edge A-->B in graph, A must before B in the order list.
    The first node in the order can be any node in the graph with no nodes direct to it.
    Find any topological order for the given graph.
    Note
    You can assume that there is at least one topological order in the graph.
    
    Example
    For graph as follow: 
    图片
    
    The topological order can be:
    
    [0, 1, 2, 3, 4, 5]
    
    or
    
    [0, 2, 3, 1, 5, 4]
    
    or
    
    ....
    
    
    
    Challenge
    Can you do it in both BFS and DFS?

    这道题参考了网上一些很好的思路:

    method1:  Record the pre nodes of every node, then find out a node without pre node in each iteration and delete this node from unvisited set, add this node to result.

     1 /**
     2  * Definition for Directed graph.
     3  * class DirectedGraphNode {
     4  *     int label;
     5  *     ArrayList<DirectedGraphNode> neighbors;
     6  *     DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); }
     7  * };
     8  */
     9 public class Solution {
    10     /**
    11      * @param graph: A list of Directed graph node
    12      * @return: Any topological order for the given graph.
    13      */    
    14     public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
    15         // write your code here
    16         ArrayList<DirectedGraphNode> res = new ArrayList<DirectedGraphNode>();
    17         if (graph.size() == 0) return res;
    18         HashMap<DirectedGraphNode, Set<DirectedGraphNode>> map = new HashMap<DirectedGraphNode, Set<DirectedGraphNode>>();
    19         for (DirectedGraphNode each : graph) {
    20             map.put(each, new HashSet<DirectedGraphNode>());
    21         }
    22         for (DirectedGraphNode each : graph) {
    23             for (int i=0; i<each.neighbors.size(); i++) {
    24                 map.get(each.neighbors.get(i)).add(each);
    25             }
    26         }
    27         while (graph.size() > 0) {
    28             int index = 0;
    29             while (index < graph.size()) {
    30                 DirectedGraphNode cur = graph.get(index);
    31                 if (map.get(cur).size() == 0) {
    32                     //add the node to our result
    33                     //remove the node from the graph
    34                     res.add(cur);
    35                     graph.remove(index);
    36                     for (DirectedGraphNode elem : graph) {
    37                         if (map.get(elem).contains(cur)) {
    38                             map.get(elem).remove(cur);
    39                         }
    40                     }
    41                 }
    42                 else index++;
    43             }
    44         }
    45         return res;
    46     }
    47 }

    method2: DFS: use a recursive method, randomly pick up an unmakred node, before adding it into result list, recursively visite all its neighbors and add its neighbors into list first. In this way, we guarantee that all the nodes belong to some node's post nodes will be added to the result list first.

    To be more specific, we can modify DFS to find Topological Sorting of a graph. In DFS, we start from a vertex, we first print it and then recursively call DFS for its adjacent vertices. In topological sorting, we don’t print the vertex immediately, we first recursively call topological sorting for all its adjacent vertices, then print the current vertex. In this way, we ensure a node's neighbor nodes are always added before the node itself.

     1 /**
     2  * Definition for Directed graph.
     3  * class DirectedGraphNode {
     4  *     int label;
     5  *     ArrayList<DirectedGraphNode> neighbors;
     6  *     DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); }
     7  * };
     8  */
     9 public class Solution {
    10     /**
    11      * @param graph: A list of Directed graph node
    12      * @return: Any topological order for the given graph.
    13      */    
    14     public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
    15         // write your code here
    16         ArrayList<DirectedGraphNode> res= new ArrayList<DirectedGraphNode>();
    17         if (graph.size() == 0) return res;
    18         HashMap<DirectedGraphNode, Integer> status = new HashMap<DirectedGraphNode, Integer>();
    19         for (DirectedGraphNode elem : graph) {
    20             status.put(elem, 0);
    21         }
    22         ArrayList<DirectedGraphNode> templist = new ArrayList<DirectedGraphNode>();
    23         templist.add(null);
    24         while (hasUnvisited(graph, status, templist)) {
    25             DirectedGraphNode cur = templist.get(0);
    26             templist.set(0, null);
    27             search(cur, status, res);
    28         }
    29         return res;
    30     }
    31     
    32     public boolean hasUnvisited(ArrayList<DirectedGraphNode> graph, HashMap<DirectedGraphNode, Integer> status, ArrayList<DirectedGraphNode> templist) {
    33         for (DirectedGraphNode elem : graph) {
    34             if (status.get(elem) == 0) {
    35                 templist.set(0, elem);
    36                 return true;
    37             }
    38         }
    39         return false;
    40     }
    41     
    42     public void search(DirectedGraphNode cur, HashMap<DirectedGraphNode, Integer> status, ArrayList<DirectedGraphNode> res) {
    43         if (status.get(cur) == 1) System.out.println("not a DAG");
    44         if (status.get(cur) == 2) return;
    45         status.put(cur, 1);
    46         for (DirectedGraphNode neigh : cur.neighbors) {
    47             search(neigh, status, res);
    48         }
    49         status.put(cur, 2);
    50         res.add(0, cur);
    51     }
    52 }
  • 相关阅读:
    无线安全
    下载安装Emacs和基本配置--待更新中
    uv-pv-vv的区别
    tesseract安装及问题处理
    POJ 2187 Beauty Contest【凸包周长】
    POJ 1113 Wall【凸包周长】
    POJ 2187 Beauty Contest【旋转卡壳求凸包直径】
    POJ 2031 Building a Space Station【经典最小生成树】
    URAL 1181 Cutting a Painted Polygon【递归+分治】
    POJ 1845-Sumdiv【经典数学题目---求因子和】
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/4431722.html
Copyright © 2011-2022 走看看