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📜  在有向图中查找每个顶点的依赖关系

📅  最后修改于: 2021-10-25 03:18:44             🧑  作者: Mango

给定一个包含N个顶点和M 个边的有向图,任务是找到图中每个顶点的所有依赖关系以及具有最小依赖关系的顶点。

例子:

方法:思路是使用深度优先搜索(DFS)来解决这个问题。

  • 获取有向图作为输入。
  • 对图执行 DFS 并探索图的所有节点。
  • 在探索节点的邻居时,将计数加 1,最后返回表示依赖项数量的计数。
  • 最后,找到依赖项数最少的节点。

下面是上述方法的实现:

CPP
// C++ program to find the
// dependency of each node
 
#include 
using namespace std;
 
// Defining the graph
class Graph {
 
    // Variable to store the
    // number of vertices
    int V;
 
    // Adjacency list
    list* adjList;
 
    // Initializing the graph
public:
    Graph(int v)
    {
        V = v;
        adjList = new list[V];
    }
 
    // Adding edges
    void addEdge(int u, int v,
                 bool bidir = true)
    {
        adjList[u].push_back(v);
        if (bidir) {
            adjList[u].push_back(v);
        }
    }
 
    // Performing DFS on each node
    int dfs(int src)
    {
        // Map is used to mark
        // the current node as visited
        map visited;
        vector dependent;
        int count = 0;
 
        stack s;
 
        // Push the current vertex
        // to the stack which
        // stores the result
        s.push(src);
 
        visited[src] = true;
 
        // Traverse through the vertices
        // until the stack is empty
        while (!s.empty()) {
            int n = s.top();
            s.pop();
 
            // Recur for all the vertices
            // adjacent to this vertex
            for (auto i : adjList[n]) {
 
                // If the vertices are
                // not visited
                if (!visited[i]) {
                    dependent.push_back(i + 1);
                    count++;
 
                    // Mark the vertex as
                    // visited
                    visited[i] = true;
 
                    // Push the current vertex to
                    // the stack which stores
                    // the result
                    s.push(i);
                }
            }
        }
 
        // If the vertex has 0 dependency
        if (!count) {
            cout << "Vertex " << src + 1
                 << " is not dependent on any vertex.\n";
            return count;
        }
 
        cout << "Vertex " << src + 1 << " dependency ";
        for (auto i : dependent) {
            cout << "-> " << i;
        }
        cout << "\n";
        return count;
    }
};
 
// Function to find the
// dependency of each node
void operations(int arr[][2],
                int n, int m)
{
    // Creating a new graph
    Graph g(n);
 
    for (int i = 0; i < m; i++) {
        g.addEdge(arr[i][0],
                  arr[i][1], false);
    }
 
    int ans = INT_MAX;
    int node = 0;
 
    // Iterating through the graph
    for (int i = 0; i < n; i++) {
        int c = g.dfs(i);
 
        // Finding the node with
        // minimum number of
        // dependency
        if (c < ans) {
            ans = c;
            node = i + 1;
        }
    }
    cout << "Node " << node
         << "has minimum dependency of "
         << ans;
}
 
// Driver code
int main()
{
    int n, m;
 
    n = 6, m = 6;
 
    // Defining the edges of the
    // graph
    int arr[][2] = { { 0, 1 },
                     { 0, 2 },
                     { 2, 3 },
                     { 4, 5 },
                     { 3, 4 },
                     { 1, 5 } };
 
    operations(arr, n, m);
 
    return 0;
}


Python3
# Python3 program to find the
# dependency of each node
 
# Adding edges
def addEdge(u, v, bidir = True):
    global adjList
    adjList[u].append(v)
    if (bidir):
        adjList[u].append(v)
 
# Performing DFS on each node
def dfs(src):
    global adjList, V
     
    # Map is used to mark
    # the current node as visited
    visited = [False for i in range(V+1)]
    dependent = []
    count = 0
    s = []
 
    # Push the current vertex
    # to the stack which
    # stores the result
    s.append(src)
    visited[src] = True
 
    # Traverse through the vertices
    # until the stack is empty
    while (len(s) > 0):
        n = s[-1]
        del s[-1]
 
        # Recur for all the vertices
        # adjacent to this vertex
        for i in adjList[n]:
 
            # If the vertices are
            # not visited
            if (not visited[i]):
                dependent.append(i + 1)
                count += 1
 
                # Mark the vertex as
                # visited
                visited[i] = True
 
                # Push the current vertex to
                # the stack which stores
                # the result
                s.append(i)
 
    # If the vertex has 0 dependency
    if (not count):
        print("Vertex ", src + 1,
              " is not dependent on any vertex.")
        return count
 
    print("Vertex ",src + 1," dependency ",end="")
    for i in dependent:
        print("-> ", i, end = "")
    print()
    return count
 
# Function to find the
# dependency of each node
def operations(arr, n, m):
   
    # Creating a new graph
    global adjList
    for i in range(m):
        addEdge(arr[i][0], arr[i][1], False)
    ans = 10**18
    node = 0
 
    # Iterating through the graph
    for i in range(n):
        c = dfs(i)
 
        # Finding the node with
        # minimum number of
        # dependency
        if (c < ans):
            ans = c
            node = i + 1
    print("Node", node, "has minimum dependency of ", ans)
 
# Driver code
if __name__ == '__main__':
    V = 6
    adjList = [[] for i in range(V+1)]
    n, m = 6, 6
 
 
    # Defining the edges of the
    # graph
    arr = [ [ 0, 1 ],
             [ 0, 2 ],
             [ 2, 3 ],
             [ 4, 5 ],
             [ 3, 4 ],
             [ 1, 5 ] ]
 
    operations(arr, n, m)
 
    # This code is contributed by mohit kumar 29.


输出:
Vertex 1 dependency -> 2-> 3-> 4-> 5-> 6
Vertex 2 dependency -> 6
Vertex 3 dependency -> 4-> 5-> 6
Vertex 4 dependency -> 5-> 6
Vertex 5 dependency -> 6
Vertex 6 is not dependent on any vertex.
Node 6has minimum dependency of 0

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