Shortest distance between given nodes in a bidirectional weighted graph by removing any K edges
Given a positive integer K and a weighted undirected connected graph of N nodes and E edges as an array Edges[] of the type {u, v, W} representing the edges between Node u and Node v having weight W, the task is to find the shortest distance between the two given nodes S and D after reducing the cost of at most K edges to 0.
Examples:
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Input: N = 5, K = 1, Edges[][] = {{0, 1, 1}, {0, 4, 1}, {1, 2, 2}, {2, 3, 4}, {4, 3, 7}}, s = 0, d = 3
Output: 1
Explanation:
Below is the graph for the given test case:
There are 2 possible routes between 0 and 3 viz. {0->1->2->3} and {0->4->3}
after reducing the distance of edge 4->3 to zero, the second route becomes 0->(4, 3) and hence the minimum distance is 1.Input: N = 5, K = 2, Edges[][] = {{0, 1, 2}, {0, 2, 3}, {2, 1, 2}, {2, 3, 1}, {3, 1, 2}, {3, 4, 3}, {4, 2, 4}}, s = 0, d = 3
Ouput: 2
Approach: The given problem can be solved using DFS Traversal and storing all possible paths between the two given nodes. Follow the steps below to solve the given problem:
- Initialize a variable, say minimumCost as INT_MAX that stores the resultant shortest distance.
- Traverse all paths from node S to node D in the graph using DFS Traversal and store all the edge weights from Node S to D obtained in a vector of vectors, say edgesPath[].
- After the above steps, sort each vector stored in edgesPath[] in decreasing order.
- Traverse the vector of vectors edgesPath[] for each vector, say A[], perform the following steps:
- Find the sum of the first K largest edges in A[].
- Update the value of minimiumCost to the minimum of the current (totalSum – sum) and mininmumCost.
- After completing the above steps, print the value of minimumCost as the result.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std; // Function to get all the possible// paths from the source to destinationvoid dfs_all(int n, int s, int d, vector<vector<pair<int, int> > >& graph, vector<bool>& vis, vector<vector<int> >& edge_path, vector<int>& temp_edge){ // One possible path, reached node D if (s == d) { edge_path.push_back(temp_edge); return; } // Mark node s as visited vis[s] = true; // Calculate number of edges with // node s as connection int edges_in_a = graph[s].size(); // Traverse all the connections // of node s for (int i = 0; i < edges_in_a; i++) { // If the connected node // isn't visited if (!vis[graph[s][i].first]) { // Push back edge value // in temp_edge temp_edge.push_back( graph[s][i].second); // Call DFS function recursively dfs_all(n, graph[s][i].first, d, graph, vis, edge_path, temp_edge); // Pop back last added edge temp_edge.pop_back(); } } // Mark s as unvisited for more // possible paths vis[s] = false;} // Function to find the minimum sum of// edges from source to destination// after reducing at most K cost to 0int getDistance( vector<vector<int> >& edge_path, int k){ // Store the shortestDistance int shortestDistance = INT_MAX; // If edge_path vector is empty, // means no path exist if (edge_path.empty()) return -1; // Traverse all the vector in // the edge_path for (auto x : edge_path) { // Base Case if (k == x.size()) return 0; // lets sort the vector in // decreasing order sort(x.begin(), x.end(), greater<int>()); // Find the sum of all the nodes int sum = 0; // Find the sum of k largest nodes int ksum = 0; for (int i = 0; i < x.size(); i++) { sum += x[i]; if (i < k) ksum += x[i]; } // If the given shortestDistance // is shortest, then update the // shortestDistance shortestDistance = min(sum - ksum, shortestDistance); } // Return the shortestDistance return shortestDistance;} // Function to find the minimum sum of// weight of edges among all paths from// source to destination after reducing// at most K cost to 0int solve( vector<vector<pair<int, int> > > graph, int n, int k, int src, int dest){ // Stores all the vectors of edges for // every path traversed in DFS call vector<vector<int> > edge_path; // Store the edges of particular path vector<int> temp_edge; // Boolean visited vector vector<bool> vis(n, false); // DFS Call dfs_all(n, src, dest, graph, vis, edge_path, temp_edge); return getDistance(edge_path, k);} // Driver Codeint main(){ int n = 5, e = 5, k = 1; vector<vector<pair<int, int> > > graph(n); // Given Adjacency List graph[0].push_back(make_pair(1, 1)); graph[1].push_back(make_pair(0, 1)); graph[0].push_back(make_pair(4, 1)); graph[4].push_back(make_pair(0, 1)); graph[1].push_back(make_pair(2, 2)); graph[2].push_back(make_pair(1, 2)); graph[2].push_back(make_pair(3, 4)); graph[3].push_back(make_pair(2, 4)); graph[4].push_back(make_pair(3, 7)); graph[3].push_back(make_pair(4, 7)); int a = 0, b = 3; cout << solve(graph, n, k, a, b); return 0;} |
1
Time Complexity: O((N*log N)NN)
Auxiliary Space: O(N2)
Efficient Approach: The above approach can also be optimized at the step where sorting is performed after finding all possible paths. Instead of sorting, the idea is to use MinHeap to calculate the sum of K largest weights in the graph to reduce the time complexity to O(N*log K) for that steps.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std; // Function to get all the possible// paths from the source to destinationvoid dfs_all(int n, int s, int d, vector<vector<pair<int, int> > >& graph, vector<bool>& vis, vector<vector<int> >& edge_path, vector<int>& temp_edge){ // One possible path, reached node D if (s == d) { edge_path.push_back(temp_edge); return; } // Mark node s as visited vis[s] = true; // Calculate number of edges with // node s as connection int edges_in_a = graph[s].size(); // Traverse all the connections // of node s for (int i = 0; i < edges_in_a; i++) { // If the connected node // isn't visited if (!vis[graph[s][i].first]) { // Push back edge value // in temp_edge temp_edge.push_back( graph[s][i].second); // Call DFS function recursively dfs_all(n, graph[s][i].first, d, graph, vis, edge_path, temp_edge); // Pop back last added edge temp_edge.pop_back(); } } // Mark s as unvisited for more // possible paths vis[s] = false;} // Function to find the minimum sum of// edges from source to destination// after reducing at most K cost to 0int getDistance( vector<vector<int> >& edge_path, int k){ int shortestDistance = INT_MAX; // If edge_path vector is empty, // means no path exist if (edge_path.empty()) return -1; // Traverse all the vector in // the edge_path for (auto x : edge_path) { if (k == x.size()) return 0; // Use heap to store the array priority_queue<int, vector<int>, greater<int> > minHeap; // Find the sum of all the nodes int sum = 0; // Find the sum of k largest nodes int ksum = 0; // Find the largest K edges using // minHeap for (int i = 0; i < x.size(); i++) { sum += x[i]; ksum += x[i]; // Pushing edge in MinHeap minHeap.push(x[i]); // If heap size is K if (minHeap.size() > k) { ksum -= minHeap.top(); minHeap.pop(); } } // If the shortestDistance is // smallest, then update the // shortestDistance shortestDistance = min(sum - ksum, shortestDistance); } // Return the shortestDistance return shortestDistance;} // Function to find the minimum sum of// weight of edges among all paths from// source to destination after reducing// at most K cost to 0int solve( vector<vector<pair<int, int> > > graph, int n, int k, int src, int dest){ // Stores all the vectors of edges for // every path traversed in DFS call vector<vector<int> > edge_path; // Store the edges of particular path vector<int> temp_edge; // Boolean visited vector vector<bool> vis(n, false); // DFS Call dfs_all(n, src, dest, graph, vis, edge_path, temp_edge); return getDistance(edge_path, k);} // Driver Codeint main(){ int n = 5, e = 5, k = 1; vector<vector<pair<int, int> > > graph(n); // Given Adjacency List graph[0].push_back(make_pair(1, 1)); graph[1].push_back(make_pair(0, 1)); graph[0].push_back(make_pair(4, 1)); graph[4].push_back(make_pair(0, 1)); graph[1].push_back(make_pair(2, 2)); graph[2].push_back(make_pair(1, 2)); graph[2].push_back(make_pair(3, 4)); graph[3].push_back(make_pair(2, 4)); graph[4].push_back(make_pair(3, 7)); graph[3].push_back(make_pair(4, 7)); int a = 0, b = 3; cout << solve(graph, n, k, a, b); return 0;} |
1
Time Complexity: O((N*log K)NN)
Auxiliary Space: O(N2)
