C++ 程序找到给定图 G 的传递闭包


如果给定有向图,则确定对于给定图中的所有顶点对 (i, j),顶点 j 是否可以从另一个顶点 i 到达。可达意为从顶点 i 到 j 有路径。这种可达性矩阵称为图的传递闭包。沃肖尔算法通常用于查找给定图 G 的传递闭包。这里有一个 C++ 程序来实现此算法。

算法

Begin
   1. Take maximum number of nodes as input.
   2. For Label the nodes as a, b, c…..
   3. To check if there any edge present between the nodes
   construct a for loop:
   // ASCII code of a is 97
   for i = 97 to (97 + n_nodes)-1
      for j = 97 to (97 + n_nodes)-1
         If edge is present do,
            adj[i - 97][j - 97] = 1
         else
            adj[i - 97][j - 97] = 0
      End loop
   End loop.
   4. To print the transitive closure of graph:
   for i = 0 to n_ nodes-1
      c = 97 + i
   End loop.
   for i = 0 to n_nodes-1
      c = 97 + i
      for j = 0 to n_nodes-1
         Print adj[I][j]
      End loop
   End loop
End

示例

#include<iostream>
using namespace std;
const int n_nodes = 20;
int main() {
   int n_nodes, k, n;
   char i, j, res, c;
   int adj[10][10], path[10][10];
   cout << "\n\tMaximum number of nodes in the graph :";
   cin >> n;
   n_nodes = n;
   cout << "\nEnter 'y'for 'YES' and 'n' for 'NO' \n";
   for (i = 97; i < 97 + n_nodes; i++)
      for (j = 97; j < 97 + n_nodes; j++) {
         cout << "\n\tIs there an edge from " << i << " to " << j << " ? ";
         cin >> res;
         if (res == 'y')
            adj[i - 97][j - 97] = 1;
         else
            adj[i - 97][j - 97] = 0;
      }
      cout << "\nTransitive Closure of the Graph:\n";
      cout << "\n\t\t\t ";
      for (i = 0; i < n_nodes; i++) {
         c = 97 + i;
         cout << c << " ";
      }
      cout << "\n\n";
      for (int i = 0; i < n_nodes; i++) {
         c = 97 + i;
         cout << "\t\t\t" << c << " ";
      for (int j = 0; j < n_nodes; j++)
         cout << adj[i][j] << " ";
         cout << "\n";
      }
      return 0;
}

输出

Maximum number of nodes in the graph :4
Enter 'y'for 'YES' and 'n' for 'NO'
Is there an edge from a to a ? y
Is there an edge from a to b ?y
Is there an edge from a to c ? n
Is there an edge from a to d ? n
Is there an edge from b to a ? y
Is there an edge from b to b ? n
Is there an edge from b to c ? y
Is there an edge from b to d ? n
Is there an edge from c to a ? y
Is there an edge from c to b ? n
Is there an edge from c to c ? n
Is there an edge from c to d ? n
Is there an edge from d to a ? y
Is there an edge from d to b ? n
Is there an edge from d to c ? y
Is there an edge from d to d ? n
Transitive Closure of the Graph:
a b c d
a 1 1 0 0
b 1 0 1 0
c 1 0 0 0
d 1 0 1 0

更新于:2019 年 7 月 30 日

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