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法律研究C++ 程序以查找图的传递闭包
如果给定了一个有向图，判断在给定的图中，对于所有顶点对 (i, j)，顶点 j 是否可以从另一个顶点 i 处到达。可到达意味着从顶点 i 到 j 有路径。这个可达矩阵被称为图的传递闭包。Warshall 算法常用于查找指定图 G 的传递闭包。下面是一个 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 make 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 << "\n\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：30- 七月-2019
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