如何在 Java 中生成给定边数的随机有向无环图？
原文:https://www . geesforgeks . org/如何在 java 中为给定的边数生成随机有向无环图/
一个有向无环图是一个没有有向圈的有向图。在有向图中，边是相连的，因此每条边都只有一个方向。一个有向无环图表示这个图不是循环的，或者说不可能从图中的一个点开始遍历整个图。每条边都从较早的边指向较晚的边。
为给定数量的边生成一个随机的有向无环图。
有向无环图
示例:
Input:
Enter the number of Edges :
20
Output:
The Generated Random Graph is :
1 -> { Isolated Vertex! }
2 -> { Isolated Vertex! }
3 -> { 18 }
4 -> { 5 }
5 -> { 16 8 }
6 -> { Isolated Vertex! }
7 -> { Isolated Vertex! }
8 -> { }
9 -> { Isolated Vertex! }
10 -> { Isolated Vertex! }
11 -> { Isolated Vertex! }
12 -> { }
13 -> { Isolated Vertex! }
14 -> { 18 }
15 -> { Isolated Vertex! }
16 -> { }
17 -> { 19 3 5 4 }
18 -> { }
19 -> { }
20 -> { 12 }
Input:
Enter the number of Edges :
30
Output:
The Generated Random Graph is :
1 -> { 12 8 7 16 5 11 }
2 -> { 16 }
3 -> { }
4 -> { 10 }
5 -> { }
6 -> { 7 }
7 -> { 5 }
8 -> { 7 12 20 }
9 -> { 16 12 }
10 -> { 3 }
11 -> { 17 14 }
12 -> { 4 3 }
13 -> { 12 5 }
14 -> { 15 17 }
15 -> { }
16 -> { 20 }
17 -> { 20 13 }
18 -> { }
19 -> { 12 11 }
20 -> { 18 }
进场:
- 输入随机有向无环图的边数。
- 在两个随机顶点之间建立一个连接，并检查是否由于这条边产生了任何循环。
- 如果发现任何循环，则丢弃该边，并再次生成随机顶点对。
实施:
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to Generate a Random Directed
// Acyclic Graph for a Given Number of Edges
import java.io.*;
import java.util.*;
import java.util.Random;
public class RandomDAG {
// The maximum number of vertex for the random graph
static int maxVertex = 20;
// Function to check for cycle, upon addition of a new
// edge in the graph
public static boolean checkAcyclic(int[][] edge, int ed,
boolean[] check, int v)
{
int i;
boolean value;
// If the current vertex is visited already, then
// the graph contains cycle
if (check[v] == true)
return false;
else {
check[v] = true;
// For each vertex, go for all the vertex
// connected to it
for (i = ed; i >= 0; i--) {
if (edge[i][0] == v)
return checkAcyclic(edge, ed, check, edge[i][1]);
}
}
// In case, if the path ends then reassign the
// vertexes visited in that path to false again
check[v] = false;
if (i == 0)
return true;
return true;
}
// Function to generate random graph
public static void generateRandomGraphs(int e)
{
int i = 0, j = 0, count = 0;
int[][] edge = new int[e][2];
boolean[] check = new boolean[21];
Random rand = new Random();
// Build a connection between two random vertex
while (i < e) {
edge[i][0] = rand.nextInt(maxVertex) + 1;
edge[i][1] = rand.nextInt(maxVertex) + 1;
for (j = 1; j <= 20; j++)
check[j] = false;
if (checkAcyclic(edge, i, check, edge[i][0]) == true)
i++;
// Check for cycle and if found discard this
// edge and generate random vertex pair again
}
System.out.println("The Generated Random Graph is :");
// Print the Graph
for (i = 0; i < maxVertex; i++) {
count = 0;
System.out.print((i + 1) + " -> { ");
for (j = 0; j < e; j++) {
if (edge[j][0] == i + 1) {
System.out.print(edge[j][1] + " ");
count++;
}
else if (edge[j][1] == i + 1) {
count++;
}
else if (j == e - 1 && count == 0)
System.out.print("Isolated Vertex!");
}
System.out.print(" }\n");
}
}
public static void main(String args[]) throws Exception
{
int e = 4;
System.out.println("Enter the number of Edges :"+ e);
// Function to generate a Random Directed Acyclic
// Graph
generateRandomGraphs(e);
}
}
Output
Enter the number of Edges :4
The Generated Random Graph is :
1 -> { Isolated Vertex! }
2 -> { 10 }
3 -> { }
4 -> { Isolated Vertex! }
5 -> { }
6 -> { 11 }
7 -> { Isolated Vertex! }
8 -> { Isolated Vertex! }
9 -> { Isolated Vertex! }
10 -> { 5 }
11 -> { }
12 -> { Isolated Vertex! }
13 -> { Isolated Vertex! }
14 -> { Isolated Vertex! }
15 -> { 3 }
16 -> { Isolated Vertex! }
17 -> { Isolated Vertex! }
18 -> { Isolated Vertex! }
19 -> { Isolated Vertex! }
20 -> { Isolated Vertex! }
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