Given a positive integer N. Consider a matrix of N X N. No cell can be accessible from any other cell, except the given pair cell in the form of (x1, y1), (x2, y2) i.e there is a path (accessible) between (x2, y2) to (x1, y1). The task is to find the count of pairs (a1, b1), (a2, b2) such that cell (a2, b2) is not accessible from (a1, b1).
Examples:
Input : N = 2 Allowed path 1: (1, 1) (1, 2) Allowed path 2: (1, 2) (2, 2) Output : 6 Cell (2, 1) is not accessible from any cell and no cell is accessible from it.(1, 1) - (2, 1) (1, 2) - (2, 1) (2, 2) - (2, 1) (2, 1) - (1, 1) (2, 1) - (1, 2) (2, 1) - (2, 2)
Consider each cell as a node, numbered from 1 to N*N. Each cell (x, y) can be map to number using (x – 1)*N + y. Now, consider each given allowed path as an edge between nodes. This will form a disjoint set of the connected component. Now, using Depth First Traversal or Breadth First Traversal, we can easily find the number of nodes or size of a connected component, say x. Now, count of non-accessible paths are x*(N*N – x). This way we can find non-accessible paths for each connected path.
Below is implementation of this approach:
C++
// C++ program to count number of pair of positions // in matrix which are not accessible #include<bits/stdc++.h> using namespace std; // Counts number of vertices connected in a component // containing x. Stores the count in k. void dfs(vector<int> graph[], bool visited[], int x, int *k) { for (int i = 0; i < graph[x].size(); i++) { if (!visited[graph[x][i]]) { // Incrementing the number of node in // a connected component. (*k)++; visited[graph[x][i]] = true; dfs(graph, visited, graph[x][i], k); } } } // Return the number of count of non-accessible cells. int countNonAccessible(vector<int> graph[], int N) { bool visited[N*N + N]; memset(visited, false, sizeof(visited)); int ans = 0; for (int i = 1; i <= N*N; i++) { if (!visited[i]) { visited[i] = true; // Initialize count of connected // vertices found by DFS starting // from i. int k = 1; dfs(graph, visited, i, &k); // Update result ans += k * (N*N - k); } } return ans; } // Inserting the edge between edge. void insertpath(vector<int> graph[], int N, int x1, int y1, int x2, int y2) { // Mapping the cell coordinate into node number. int a = (x1 - 1) * N + y1; int b = (x2 - 1) * N + y2; // Inserting the edge. graph[a].push_back(b); graph[b].push_back(a); } // Driven Program int main() { int N = 2; vector<int> graph[N*N + 1]; insertpath(graph, N, 1, 1, 1, 2); insertpath(graph, N, 1, 2, 2, 2); cout << countNonAccessible(graph, N) << endl; return 0; } |
Python3
# Python3 program to count number of pair of # positions in matrix which are not accessible # Counts number of vertices connected in a # component containing x. Stores the count in k. def dfs(graph,visited, x, k): for i in range(len(graph[x])): if (not visited[graph[x][i]]): # Incrementing the number of node # in a connected component. k[0] += 1 visited[graph[x][i]] = True dfs(graph, visited, graph[x][i], k) # Return the number of count of # non-accessible cells. def countNonAccessible(graph, N): visited = [False] * (N * N + N) ans = 0 for i in range(1, N * N + 1): if (not visited[i]): visited[i] = True # Initialize count of connected # vertices found by DFS starting # from i. k = [1] dfs(graph, visited, i, k) # Update result ans += k[0] * (N * N - k[0]) return ans # Inserting the edge between edge. def insertpath(graph, N, x1, y1, x2, y2): # Mapping the cell coordinate # into node number. a = (x1 - 1) * N + y1 b = (x2 - 1) * N + y2 # Inserting the edge. graph[a].append(b) graph[b].append(a) # Driver Code if __name__ == '__main__': N = 2 graph = [[] for i in range(N*N + 1)] insertpath(graph, N, 1, 1, 1, 2) insertpath(graph, N, 1, 2, 2, 2) print(countNonAccessible(graph, N)) # This code is contributed by PranchalK |
Output:
6
Time Complexity : O(N * N).
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