【发布时间】：2012-10-14 19:36:59
【问题描述】：
我正在研究一种算法，以在网格中找到一组不相交的路径 给定点对.. 对这些对像这样： (9,4) 和 (12,13)
输出应该是这样的：
9,10,11,7,3,4
13,14,15,16,12
如果无法路由所有路径，则打印“Blocked”
首先我搜索了一个已经创建的算法来找到 2 之间的所有简单路径 图形或网格中的点。我通过@Casey Watson 和@svick here..找到了这个。 它的效果非常好，但仅适用于小图。
我将其转换为 C#.NET 并对其进行了一些改进，以便能够找到 最大长度 X。并以此为基础构建我的总算法。
我构建的那个在小图中工作得很好。 这是 8x8 网格中的 9 对路线。
但是在像 16x16 甚至我打算做的最后一个是 16x16x2 的 3D 模型这样较大的模型上需要很长时间 像这样
该算法被开发为深度优先搜索RECURSIVE算法，但它 花费大量时间向用户返回价值。所以我决定将其转换为循环而不是递归调用，以便我可以从 .NET 中的 yield return 功能中受益 但它仍然没有更好的帮助。
算法的循环版本在不到一秒的时间内找到一对点的路线，但递归版本需要 90 多秒。
当我尝试使用 2 对时，循环版本大约需要 342 秒，但递归版本大约需要 200 秒..
所以我不知道哪个更快..！？递归或循环之一..
我真的很想知道最好的方法..
注意：节点编号中的第一个数字确定层（从1开始）..
这里是代码
using System;
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
namespace AlgorithmTest
{
struct Connection
{
public int FirstNode;
public int SecondNode;
public Connection(int N1,int N2)
{
FirstNode = N1;
SecondNode = N2;
}
}
enum Algorithm
{ Recursion, Loops }
public class Search
{
private const int MAX = 15;
private const int Width = 16;
private const int Length = 16;
private const int Height = 2;
private static void Main(string[] args)
{
var graph = new Graph();
var str = new int[Height,Length, Width];
var level = ((int)Math.Pow(10, (Length * Width).ToString().Length) >= 100) ? (int)Math.Pow(10, (Length * Width).ToString().Length) : 100;
for (var i = 0; i < Height; i++)
{
int num = 0;
for (var j = 0; j < Length; j++)
for (var k = 0; k < Width; k++)
{
str[i, j, k] = ++num + level;
}
level += level;
}
for (var i = 0; i < Height; i++)
{
for (var j = 0; j < Length; j++)
{
for (var k = 0; k < Width; k++)
{
if (i < Height - 1) graph.addEdge(str[i, j, k], str[i + 1, j, k]);
if (i > 0) graph.addEdge(str[i, j, k], str[i - 1, j, k]);
if (k < Width - 1) graph.addEdge(str[i, j, k], str[i, j, k + 1]);
if (k > 0) graph.addEdge(str[i, j, k], str[i, j, k - 1]);
if (j < Length - 1) graph.addEdge(str[i, j, k], str[i, j + 1, k]);
if (j > 0) graph.addEdge(str[i, j, k], str[i, j - 1, k]);
}
}
}
var wt = new Stopwatch();
wt.Start();
var connectedNodes = new List<Connection>()
{
new Connection(1030, 1005),
// new Connection(1002, 1044),
// new Connection(1015, 1064),
// new Connection(1041, 1038),
// new Connection(1009, 1027),
// new Connection(1025, 1018),
// new Connection(1037, 1054),
// new Connection(1049, 1060),
// new Connection(1008, 1031),
// new Connection(1001, 1035),
};
wt.Start();
Console.WriteLine("Using Loops:");
Console.WriteLine();
var allPaths = new Search().FindAllPaths(connectedNodes, graph, MAX, Algorithm.Loops);
wt.Stop();
foreach (var path in allPaths)
{
PrintPath(path);
}
Console.WriteLine("Total Seconds: " + wt.Elapsed.TotalSeconds + ", Number of paths: " + allPaths.Count());
Console.WriteLine("***************************************************************************************************");
Console.WriteLine("Using Recursion:");
Console.WriteLine();
wt.Reset();
wt.Start();
allPaths = new Search().FindAllPaths(connectedNodes, graph, MAX, Algorithm.Recursion);
wt.Stop();
foreach (var path in allPaths)
{
PrintPath(path);
}
Console.WriteLine("Total Seconds: " + wt.Elapsed.TotalSeconds + ", Number of paths: " + allPaths.Count());
Console.WriteLine();
}
private IEnumerable<List<int>> FindAllPaths(List<Connection> connectedNodes, Graph graph, int max, Algorithm algorithm)
{
var paths=new Stack<List<int>>();
var blocked=new List<int>();
for (var i = 0; i < connectedNodes.Count; i++)
{
if (!blocked.Contains(connectedNodes[i].FirstNode)) blocked.Add(connectedNodes[i].FirstNode);
if (!blocked.Contains(connectedNodes[i].SecondNode)) blocked.Add(connectedNodes[i].SecondNode);
}
if (algorithm == Algorithm.Recursion)
{
if (FindAllPaths(connectedNodes, 0, max, graph, paths, blocked))
{
Console.WriteLine("BLOCKED");
return new List<List<int>>();
}
}
else if(algorithm==Algorithm.Loops)
{
if (!FindAllPaths2(connectedNodes, 0, max, graph, paths, blocked))
{
Console.WriteLine("BLOCKED");
return new List<List<int>>();
}
}
return paths;
}
private static bool FindAllPaths(List<Connection> connectedNodes,int order,int max, Graph graph, Stack<List<int>> allPaths, List<int> blocked)
{
if (order >= connectedNodes.Count) return false;
var paths = SearchForPaths(graph, connectedNodes[order].FirstNode, connectedNodes[order].SecondNode, max, blocked);
if (paths.Count == 0) return true;
int i;
for (i = 0; i < paths.Count; i++)
{
var path = paths[i];
allPaths.Push(path);
blocked.AddRange(path);
if (!FindAllPaths(connectedNodes, order + 1,max, graph, allPaths, blocked)) break;
allPaths.Pop();
foreach (var j in path)
{
blocked.RemoveAll(num => num==j);
}
paths.RemoveAll(list => IsListsSimilar(list,path));
i--;
}
if (i == paths.Count) return true;
return false;
}
private static bool IsListsSimilar(List<int> L1,List<int> L2)
{
if (L2.Count > L1.Count) return false;
for (int i = 0; i < L2.Count - 1; i++)
{
if (L1[i] != L2[i]) return false;
}
return true;
}
private static List<List<int>> SearchForPaths(Graph graph, int start, int end, int max, List<int> blocked)
{
blocked.Remove(start);
blocked.Remove(end);
var nodePaths = new List<List<int>>();
var visited = new LinkedList<int>();
visited.AddLast(start);
DepthFirstSearch(graph, visited, end, max, blocked, nodePaths);
nodePaths = nodePaths.OrderBy(list => list.Count).ToList();
return nodePaths;
}
private static void DepthFirstSearch(Graph graph, LinkedList<int> visited, int end, int max, List<int> blocked, List<List<int>> paths)
{
var nodes = graph.adjacentNodes(visited.Last.Value);
// examine adjacent nodes
var nodeCount = blocked.Count;
for (int i = 0; i < nodeCount; i++)
{
if (visited.Contains(blocked[i])) return;
}
if (visited.Count > max) return;
nodeCount = nodes.Count;
for (var i = 0; i < nodeCount; i++)
{
if (visited.Contains(nodes[i]) || nodes[i] != end) continue;
visited.AddLast(nodes[i]);
{
paths.Add(new List<int>(visited));
}
visited.RemoveLast();
break;
}
nodeCount = nodes.Count;
for (var i = 0; i < nodeCount; i++)
{
if (visited.Contains(nodes[i]) || nodes[i] == end) continue;
visited.AddLast(nodes[i]);
DepthFirstSearch(graph, visited, end, max, blocked, paths);
visited.RemoveLast();
}
}
private static bool FindAllPaths2(List<Connection> connectedNodes, int order, int max, Graph graph, Stack<List<int>> allPaths, List<int> blocked)
{
if (order >= connectedNodes.Count) return false;
foreach (var path in SearchForPaths2(graph, connectedNodes[order].FirstNode, connectedNodes[order].SecondNode, max, blocked))
{
allPaths.Push(path);
blocked.AddRange(path);
if (!FindAllPaths2(connectedNodes, order + 1, max, graph, allPaths, blocked)) break;
allPaths.Pop();
foreach (var j in path)
{
blocked.RemoveAll(num => num == j);
}
}
return true;
}
private static IEnumerable<List<int>> SearchForPaths2(Graph graph, int start, int end, int max, List<int> blocked)
{
blocked.Remove(start);
blocked.Remove(end);
var visited = new LinkedList<int>();
visited.AddLast(start);
foreach (var VARIABLE in DepthFirstSearch(graph, visited, end, max, blocked))
{
yield return VARIABLE;
}
}
private static IEnumerable<List<int>> DepthFirstSearch(Graph graph, LinkedList<int> visited, int end, int max, List<int> blocked)
{
var nodes = graph.adjacentNodes(visited.Last.Value);
var nodeCount = blocked.Count;
for (int i = 0; i < nodeCount; i++)
{
if (visited.Contains(blocked[i])) yield break;
}
if (visited.Count > max) yield break;
nodeCount = nodes.Count;
for (var i = 0; i < nodeCount; i++)
{
if (visited.Contains(nodes[i]) || nodes[i] != end) continue;
visited.AddLast(nodes[i]);
yield return (new List<int>(visited));
visited.RemoveLast();
break;
}
nodeCount = nodes.Count;
for (var i = 0; i < nodeCount; i++)
{
if (visited.Contains(nodes[i]) || nodes[i] == end) continue;
visited.AddLast(nodes[i]);
foreach (var P in DepthFirstSearch(graph, visited, end, max, blocked))
{
yield return P;
}
visited.RemoveLast();
}
}
private static void PrintPath(List<int> visited)
{
for (int i = 0; i < visited.Count()-1; i++)
{
Console.Write(visited[i]);
Console.Write(" --> ");
}
Console.Write(visited[visited.Count() - 1]);
Console.WriteLine();
Console.WriteLine();
}
}
public class Graph
{
private readonly Dictionary<int, HashSet<int>> map = new Dictionary<int, HashSet<int>>();
public void addEdge(int node1, int node2)
{
HashSet<int> adjacent = null;
map.TryGetValue(node1, out adjacent);
if (adjacent == null)
{
adjacent = new HashSet<int>();
map.Add(node1, adjacent);
}
adjacent.Add(node2);
}
public List<int> adjacentNodes(int last)
{
HashSet<int> adjacent = null;
map.TryGetValue(last, out adjacent);
if (adjacent == null)
{
return new List<int>();
}
return new List<int>(adjacent);
}
}
}
【问题讨论】：
-
您是否尝试过其他算法，例如
A*或Dijkstra，应该可以相当快地计算出最短路径，另请参阅：codeproject.com/Articles/118015/… -
@Jos，我从来没有考虑过，因为我不想计算最短路径，我只想在网格中路由 多个 点对而不相交。但我会尝试修改 Dijkstra 算法并再试一次..
-
好的，谢谢，我第一次阅读时并没有这样理解。我认为您可以通过从网格中“删除”使用的顶点并多次重复最短路径来做到这一点。但我不确定这是否是“理想”的解决方案。无论如何，祝你的项目好运。 :)
-
我会尝试，但我会删除检查路径甚至类似的路径。但正如你所说，我认为我的问题有一个理想的解决方案，我正在等待答案.. :)
-
我不明白是什么问题。您是否在问递归是否比循环更有效，反之亦然？或者您是否要求一种有效的算法来解决非相交路径问题？
标签： c# .net performance