Given a weighted undirected graph T consisting of nodes valued [0, N – 1] and an array Edges[][3] of type {u, v, w} that denotes an edge between vertices u and v having weight w. The task is to find the sum of all pair shortest paths in the given tree.
Examples:
Input: N = 3, Edges[][] = {{0, 2, 15}, {1, 0, 90}}
Output: 210
Explanation:
Sum of weights of path between nodes 0 and 1 = 90
Sum of weights of path between nodes 0 and 2 = 15
Sum of weights of path between nodes 1 and 2 = 105
Hence, sum = 90 + 15 + 105Input: N = 4, Edges[][] = {{0, 1, 1}, {1, 2, 2}, {2, 3, 3}}
Output: 20
Explanation:
Sum of weights of path between nodes 0 and 1 = 1
Sum of weights of path between nodes 0 and 2 = 3
Sum of weights of path between nodes 0 and 3 = 6
Sum of weights of path between nodes 1 and 2 = 2
Sum of weights of path between nodes 1 and 3 = 5
Sum of weights of path between nodes 2 and 3 = 3
Hence, sum = 1 + 3 + 6 + 2 + 5 + 3 = 20.
Naive Approach: The simplest approach is to find the shortest path between every pair of vertices using the Floyd Warshall Algorithm. After precomputing the cost of the shortest path between every pair of nodes, print the sum of all the shortest paths.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;#define INF 99999// Function that performs the Floyd// Warshall to find all shortest pathsint floyd_warshall(int* graph, int V){ int dist[V][V], i, j, k; // Initialize the distance matrix for (i = 0; i < V; i++) { for (j = 0; j < V; j++) { dist[i][j] = *((graph + i * V) + j); } } for (k = 0; k < V; k++) { // Pick all vertices as // source one by one for (i = 0; i < V; i++) { // Pick all vertices as // destination for the // above picked source for (j = 0; j < V; j++) { // If vertex k is on the // shortest path from i to // j then update dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]) { dist[i][j] = dist[i][k] + dist[k][j]; } } } } // Sum the upper diagonal of the // shortest distance matrix int sum = 0; // Traverse the given dist[][] for (i = 0; i < V; i++) { for (j = i + 1; j < V; j++) { // Add the distance sum += dist[i][j]; } } // Return the final sum return sum;}// Function to generate the treeint sumOfshortestPath(int N, int E, int edges[][3]){ int g[N][N]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { g[i][j] = INF; } } // Add edges for (int i = 0; i < E; i++) { // Get source and destination // with weight int u = edges[i][0]; int v = edges[i][1]; int w = edges[i][2]; // Add the edges g[u][v] = w; g[v][u] = w; } // Perform Floyd Warshal Algorithm return floyd_warshall((int*)g, N);}// Driver codeint main(){ // Number of Vertices int N = 4; // Number of Edges int E = 3; // Given Edges with weight int Edges[][3] = { { 0, 1, 1 }, { 1, 2, 2 }, { 2, 3, 3 } }; // Function Call cout << sumOfshortestPath(N, E, Edges); return 0;} |
Java
// Java program for // the above approachclass GFG{ static final int INF = 99999;// Function that performs the Floyd// Warshall to find all shortest pathsstatic int floyd_warshall(int[][] graph, int V){ int [][]dist = new int[V][V]; int i, j, k; // Initialize the distance matrix for (i = 0; i < V; i++) { for (j = 0; j < V; j++) { dist[i][j] = graph[i][j]; } } for (k = 0; k < V; k++) { // Pick all vertices as // source one by one for (i = 0; i < V; i++) { // Pick all vertices as // destination for the // above picked source for (j = 0; j < V; j++) { // If vertex k is on the // shortest path from i to // j then update dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]) { dist[i][j] = dist[i][k] + dist[k][j]; } } } } // Sum the upper diagonal of the // shortest distance matrix int sum = 0; // Traverse the given dist[][] for (i = 0; i < V; i++) { for (j = i + 1; j < V; j++) { // Add the distance sum += dist[i][j]; } } // Return the final sum return sum;}// Function to generate the treestatic int sumOfshortestPath(int N, int E, int edges[][]){ int [][]g = new int[N][N]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { g[i][j] = INF; } } // Add edges for (int i = 0; i < E; i++) { // Get source and destination // with weight int u = edges[i][0]; int v = edges[i][1]; int w = edges[i][2]; // Add the edges g[u][v] = w; g[v][u] = w; } // Perform Floyd Warshal Algorithm return floyd_warshall(g, N);}// Driver codepublic static void main(String[] args){ // Number of Vertices int N = 4; // Number of Edges int E = 3; // Given Edges with weight int Edges[][] = {{0, 1, 1}, {1, 2, 2}, {2, 3, 3}}; // Function Call System.out.print( sumOfshortestPath(N, E, Edges));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approachINF = 99999# Function that performs the Floyd# Warshall to find all shortest pathsdef floyd_warshall(graph, V): dist = [[0 for i in range(V)] for i in range(V)] # Initialize the distance matrix for i in range(V): for j in range(V): dist[i][j] = graph[i][j] for k in range(V): # Pick all vertices as # source one by one for i in range(V): # Pick all vertices as # destination for the # above picked source for j in range(V): # If vertex k is on the # shortest path from i to # j then update dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]): dist[i][j] = dist[i][k] + dist[k][j] # Sum the upper diagonal of the # shortest distance matrix sum = 0 # Traverse the given dist[][] for i in range(V): for j in range(i + 1, V): # Add the distance sum += dist[i][j] # Return the final sum return sum# Function to generate the treedef sumOfshortestPath(N, E,edges): g = [[INF for i in range(N)] for i in range(N)] # Add edges for i in range(E): # Get source and destination # with weight u = edges[i][0] v = edges[i][1] w = edges[i][2] # Add the edges g[u][v] = w g[v][u] = w # Perform Floyd Warshal Algorithm return floyd_warshall(g, N)# Driver codeif __name__ == '__main__': # Number of Vertices N = 4 # Number of Edges E = 3 # Given Edges with weight Edges = [ [ 0, 1, 1 ], [ 1, 2, 2 ], [ 2, 3, 3 ] ] # Function Call print(sumOfshortestPath(N, E, Edges))# This code is contributed by mohit kumar 29 |
C#
// C# program for // the above approachusing System;class GFG{ static readonly int INF = 99999;// Function that performs the Floyd// Warshall to find all shortest pathsstatic int floyd_warshall(int[,] graph, int V){ int [,]dist = new int[V, V]; int i, j, k; // Initialize the distance matrix for (i = 0; i < V; i++) { for (j = 0; j < V; j++) { dist[i, j] = graph[i, j]; } } for (k = 0; k < V; k++) { // Pick all vertices as // source one by one for (i = 0; i < V; i++) { // Pick all vertices as // destination for the // above picked source for (j = 0; j < V; j++) { // If vertex k is on the // shortest path from i to // j then update dist[i,j] if (dist[i, k] + dist[k, j] < dist[i, j]) { dist[i, j] = dist[i, k] + dist[k, j]; } } } } // Sum the upper diagonal of the // shortest distance matrix int sum = 0; // Traverse the given dist[,] for (i = 0; i < V; i++) { for (j = i + 1; j < V; j++) { // Add the distance sum += dist[i, j]; } } // Return the readonly sum return sum;}// Function to generate the treestatic int sumOfshortestPath(int N, int E, int [,]edges){ int [,]g = new int[N, N]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { g[i, j] = INF; } } // Add edges for (int i = 0; i < E; i++) { // Get source and destination // with weight int u = edges[i, 0]; int v = edges[i, 1]; int w = edges[i, 2]; // Add the edges g[u, v] = w; g[v, u] = w; } // Perform Floyd Warshal Algorithm return floyd_warshall(g, N);}// Driver codepublic static void Main(String[] args){ // Number of Vertices int N = 4; // Number of Edges int E = 3; // Given Edges with weight int [,]Edges = {{0, 1, 1}, {1, 2, 2}, {2, 3, 3}}; // Function Call Console.Write(sumOfshortestPath(N, E, Edges));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for// the above approachlet INF = 99999;// Function that performs the Floyd// Warshall to find all shortest pathsfunction floyd_warshall(graph,V){ let dist = new Array(V); for(let i = 0; i < V; i++) { dist[i] = new Array(V); } let i, j, k; // Initialize the distance matrix for (i = 0; i < V; i++) { for (j = 0; j < V; j++) { dist[i][j] = graph[i][j]; } } for (k = 0; k < V; k++) { // Pick all vertices as // source one by one for (i = 0; i < V; i++) { // Pick all vertices as // destination for the // above picked source for (j = 0; j < V; j++) { // If vertex k is on the // shortest path from i to // j then update dist[i][j] if (dist[i][k] + dist[k][j] < dist[i][j]) { dist[i][j] = dist[i][k] + dist[k][j]; } } } } // Sum the upper diagonal of the // shortest distance matrix let sum = 0; // Traverse the given dist[][] for (i = 0; i < V; i++) { for (j = i + 1; j < V; j++) { // Add the distance sum += dist[i][j]; } } // Return the final sum return sum;}// Function to generate the treefunction sumOfshortestPath(N,E,edges){ let g = new Array(N); for (let i = 0; i < N; i++) { g[i] = new Array(N); for (let j = 0; j < N; j++) { g[i][j] = INF; } } // Add edges for (let i = 0; i < E; i++) { // Get source and destination // with weight let u = edges[i][0]; let v = edges[i][1]; let w = edges[i][2]; // Add the edges g[u][v] = w; g[v][u] = w; } // Perform Floyd Warshal Algorithm return floyd_warshall(g, N);}// Driver code// Number of Verticeslet N = 4;// Number of Edgeslet E = 3;// Given Edges with weightlet Edges = [[0, 1, 1], [1, 2, 2],[2, 3, 3]];// Function Calldocument.write(sumOfshortestPath(N, E, Edges));// This code is contributed by patel2127</script> |
Output
20
Time Complexity:O(N3), where N is the number of vertices.
Auxiliary Space: O(N)
Efficient Approach: The idea is to use the DFS algorithm, using the DFS, for each vertex, the cost to visit every other vertex from this vertex can be found in linear time. Follow the below steps to solve the problem:
- Traverse the nodes 0 to N – 1.
- For each node i, find the sum of the cost to visit every other vertex using DFS where the source will be node i, and let’s denote this sum by Si.
- Now, calculate S = S0 + S1 + … + SN-1. and divide S by 2 because every path is calculated twice.
- After completing the above steps, print the value of sum S obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include<bits/stdc++.h>using namespace std;// Function that performs the DFS// traversal to find cost to reach// from vertex v to other vertexesvoid dfs(int v, int p, vector<pair<int, int>> t[], int h, int ans[]){ // Traverse the Adjacency list // of u for(pair<int, int> u : t[v]) { if (u.first == p) continue; // Recursive Call dfs(u.first, v, t, h + u.second, ans); } // Update ans[v] ans[v] = h;}// Function to find the sum of// weights of all pathsint solve(int n, int edges[][3]){ // Stores the Adjacency List vector<pair<int, int>> t[n]; // Store the edges for(int i = 0; i < n - 1; i++) { t[edges[i][0]].push_back({edges[i][1], edges[i][2]}); t[edges[i][1]].push_back({edges[i][0], edges[i][2]}); } // sum is the answer int sum = 0; // Calculate sum for each vertex for(int i = 0; i < n; i++) { int ans[n]; // Perform the DFS Traversal dfs(i, -1, t, 0, ans); // Sum of distance for(int j = 0; j < n; j++) sum += ans[j]; } // Return the final sum return sum / 2;}// Driver Codeint main(){ // No of vertices int N = 4; // Given Edges int edges[][3] = { { 0, 1, 1 }, { 1, 2, 2 }, { 2, 3, 3 } }; // Function Call cout << solve(N, edges) << endl; return 0;}// This code is contributed by pratham76 |
Java
// Java program for the above approachimport java.io.*;import java.awt.*;import java.io.*;import java.util.*;@SuppressWarnings("unchecked")class GFG { // Function that performs the DFS // traversal to find cost to reach // from vertex v to other vertexes static void dfs(int v, int p, ArrayList<Point> t[], int h, int ans[]) { // Traverse the Adjacency list // of u for (Point u : t[v]) { if (u.x == p) continue; // Recursive Call dfs(u.x, v, t, h + u.y, ans); } // Update ans[v] ans[v] = h; } // Function to find the sum of // weights of all paths static int solve(int n, int edges[][]) { // Stores the Adjacency List ArrayList<Point> t[] = new ArrayList[n]; for (int i = 0; i < n; i++) t[i] = new ArrayList<>(); // Store the edges for (int i = 0; i < n - 1; i++) { t[edges[i][0]].add( new Point(edges[i][1], edges[i][2])); t[edges[i][1]].add( new Point(edges[i][0], edges[i][2])); } // sum is the answer int sum = 0; // Calculate sum for each vertex for (int i = 0; i < n; i++) { int ans[] = new int[n]; // Perform the DFS Traversal dfs(i, -1, t, 0, ans); // Sum of distance for (int j = 0; j < n; j++) sum += ans[j]; } // Return the final sum return sum / 2; } // Driver Code public static void main(String[] args) { // No of vertices int N = 4; // Given Edges int edges[][] = new int[][] { { 0, 1, 1 }, { 1, 2, 2 }, { 2, 3, 3 } }; // Function Call System.out.println(solve(N, edges)); }} |
Python3
# Python3 program for the above approach# Function that performs the DFS# traversal to find cost to reach# from vertex v to other vertexesdef dfs(v, p, t, h, ans): # Traverse the Adjacency list # of u for u in t[v]: if (u[0] == p): continue # Recursive Call dfs(u[0], v, t, h + u[1], ans) # Update ans[v] ans[v] = h# Function to find the sum of# weights of all pathsdef solve(n, edges): # Stores the Adjacency List t = [[] for i in range(n)] # Store the edges for i in range(n - 1): t[edges[i][0]].append([edges[i][1], edges[i][2]]) t[edges[i][1]].append([edges[i][0], edges[i][2]]) # sum is the answer sum = 0 # Calculate sum for each vertex for i in range(n): ans = [0 for i in range(n)] # Perform the DFS Traversal dfs(i, -1, t, 0, ans) # Sum of distance for j in range(n): sum += ans[j] # Return the final sum return sum // 2 # Driver Codeif __name__ == "__main__": # No of vertices N = 4 # Given Edges edges = [ [ 0, 1, 1 ], [ 1, 2, 2 ], [ 2, 3, 3 ] ] # Function Call print(solve(N, edges))# This code is contributed by rutvik_56 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function that performs the DFS// traversal to find cost to reach// from vertex v to other vertexesstatic void dfs(int v, int p, List<Tuple<int, int>> []t, int h, int []ans){ // Traverse the Adjacency list // of u foreach(Tuple<int, int> u in t[v]) { if (u.Item1 == p) continue; // Recursive call dfs(u.Item1, v, t, h + u.Item2, ans); } // Update ans[v] ans[v] = h;}// Function to find the sum of// weights of all pathsstatic int solve(int n, int [,]edges){ // Stores the Adjacency List List<Tuple<int, int>> []t = new List<Tuple<int, int>>[n]; for(int i = 0; i < n; i++) t[i] = new List<Tuple<int, int>>(); // Store the edges for(int i = 0; i < n - 1; i++) { t[edges[i, 0]].Add( new Tuple<int, int>(edges[i, 1], edges[i, 2])); t[edges[i, 1]].Add( new Tuple<int, int>(edges[i, 0], edges[i, 2])); } // sum is the answer int sum = 0; // Calculate sum for each vertex for(int i = 0; i < n; i++) { int []ans = new int[n]; // Perform the DFS Traversal dfs(i, -1, t, 0, ans); // Sum of distance for(int j = 0; j < n; j++) sum += ans[j]; } // Return the readonly sum return sum / 2;}// Driver Codepublic static void Main(String[] args){ // No of vertices int N = 4; // Given Edges int [,]edges = new int[,] { { 0, 1, 1 }, { 1, 2, 2 }, { 2, 3, 3 } }; // Function call Console.WriteLine(solve(N, edges));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// Function that performs the DFS// traversal to find cost to reach// from vertex v to other vertexesfunction dfs(v, p, t, h, ans){ // Traverse the Adjacency list // of u for(let u = 0; u < t[v].length; u++) { if (t[v][u][0] == p) continue; // Recursive Call dfs(t[v][u][0], v, t, h + t[v][u][1], ans); } // Update ans[v] ans[v] = h;}// Function to find the sum of// weights of all pathsfunction solve(n, edges){ // Stores the Adjacency List let t = new Array(n); for(let i = 0; i < n; i++) t[i] = []; // Store the edges for(let i = 0; i < n - 1; i++) { t[edges[i][0]].push([edges[i][1], edges[i][2]]); t[edges[i][1]].push([edges[i][0], edges[i][2]]); } // Sum is the answer let sum = 0; // Calculate sum for each vertex for(let i = 0; i < n; i++) { let ans = new Array(n); // Perform the DFS Traversal dfs(i, -1, t, 0, ans); // Sum of distance for(let j = 0; j < n; j++) sum += ans[j]; } // Return the final sum return sum / 2;}// Driver Codelet N = 4;let edges = [ [ 0, 1, 1 ], [ 1, 2, 2 ], [ 2, 3, 3 ] ]; document.write(solve(N, edges));// This code is contributed by unknown2108</script> |
Output
20
Time Complexity: O(N2), where N is the number of vertices.
Auxiliary Space: O(N)
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