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Check whether the cost of going from any node to any other node via all possible paths is same
  • Difficulty Level : Expert
  • Last Updated : 11 Jun, 2021

Given adjacency list representation of a directed graph, the task is to check whether the cost of going from any vertex to any other vertex via all possible paths is equal or not. If there is a cost c for going from vertex A to vertex B, then the cost to travel from vertex B to vertex A will be -c

Examples: 

Input: arr[][] = {{0, 2, 0, 1}, {-2, 0, 1, 0}, {0, -1, 0, -2}, {-1, 0, 2, 0}} 
Output: Yes 
Explanation: 
 

Here the cost of going from any node to any other node is equal for all possible paths. For example, if we go from 1 to 4 via (1 -> 2 -> 3 -> 4) whose cost is (2 + 1 + (-2)) i.e. 1 and via ( 1 -> 4 ) which is a reverse edge whose cost is 1. Similarly, the cost for all other paths is equal.



Input: arr[][] = {{0, 1, 0, 3, 0}, {-1, 0, 3, 0, 4}, {0, -3, 0, 1, 1}, {-3, 0, -1, 0, 0}, {0, -4, -1, 0, 0}} 
Output: No 
Explanation: 
 

For the following two paths from edge 1 to 4, (1 -> 2 -> 3 -> 4), Cost = (1 + 3 + 1) = 5 and (1 -> 4), Cost = 3. Since the costs are difference, the answer is No 
 

Approach: The idea is to maintain two arrays dis[] which maintains the distance of the paths travelled and visited[] which maintains the visited vertices. The graph is stored using 2D vector of pairs. The first value of the pair is the destination node and the second value is the cost associated with it. Now, DFS is run on the graph. The following two conditions occur for every vertex:

  1. If the next node to reach is not visited, then the dis array is updated with the value dis[current_node] + cost of the new edge found from the 2D vector i.e. current node to next node to reach and the same function is called with the unvisited same node.
  2. If the node is visited, then the distance of the next node to reach is compared with the dis[curr] + cost of edge to reach the next node. If they are equal, then the boolean variable flag is updated with true and the loop is continued for the next vertices.

Below is the implementation of the above approach:  

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > adj[100005];
 
// Initialize distance and visited array
int vis[100005] = { 0 },
    dist[100005] = { 0 },
    flg;
 
// Function to perform dfs and check
// For a given vertex If the distance
// for all the paths is equal or not
void dfs(int curr)
{
    vis[curr] = 1;
    for (int i = 0; i < adj[curr].size(); i++) {
 
        // Checking the next node to reach
        // is visited or not
        if (vis[adj[curr][i].first]) {
 
            // Case 2: comparing the distance
            if (dist[adj[curr][i].first]
                != dist[curr] + adj[curr][i].second)
                flg = 1;
        }
        else {
 
            // Case 1: Adding the distance
            // and updating the array
            dist[adj[curr][i].first] = dist[curr]
                                       + adj[curr][i].second;
 
            // Calling the function again with the
            // same node
            dfs(adj[curr][i].first);
        }
    }
}
 
// Driver code
int main()
{
    int n = 4, m = 4;
    flg = 0;
    // Creating the graph as mentioned
    // in example 1
    adj[0].push_back({ 1, 2 });
    adj[1].push_back({ 0, -2 });
    adj[1].push_back({ 2, 1 });
    adj[2].push_back({ 1, -1 });
    adj[2].push_back({ 3, -2 });
    adj[3].push_back({ 2, 2 });
    adj[3].push_back({ 0, -1 });
    adj[0].push_back({ 3, 1 });
    for (int i = 0; i < n; i++) {
        if (flg)
 
            // If for any vertex, flg is true,
            // then the distance is not equal
            break;
 
        if (!vis[i])
 
            // Calling the DFS function if
            // the vertex is not visited
            dfs(i);
    }
    if (flg)
        cout << "No" << endl;
    else
        cout << "Yes" << endl;
    return 0;
}

Java




// Java implementation of the above approach
import java.util.*;
 
class GFG {
    static class pair {
        int first, second;
 
        public pair(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }
 
    static Vector<pair>[] adj = new Vector[100005];
 
    // Initialize distance and visited array
    static int[] vis = new int[100005];
    static int[] dist = new int[100005];
    static int flg;
 
    // Function to perform dfs and check
    // For a given vertex If the distance
    // for all the paths is equal or not
    static void dfs(int curr) {
        vis[curr] = 1;
        for (int i = 0; i < adj[curr].size(); i++) {
 
            // Checking the next node to reach
            // is visited or not
            if (vis[adj[curr].get(i).first] > 0) {
 
                // Case 2: comparing the distance
                if (dist[adj[curr].get(i).first] !=
                    dist[curr] + adj[curr].get(i).second)
                    flg = 1;
            } else {
 
                // Case 1: Adding the distance
                // and updating the array
                dist[adj[curr].get(i).first] =
                    dist[curr] + adj[curr].get(i).second;
 
                // Calling the function again with the
                // same node
                dfs(adj[curr].get(i).first);
            }
        }
    }
 
    // Driver code
    public static void main(String[] args) {
        int n = 4, m = 4;
        flg = 0;
        for (int i = 0; i < adj.length; i++) {
            adj[i] = new Vector<pair>();
        }
        // Creating the graph as mentioned
        // in example 1
        adj[0].add(new pair(1, 2));
        adj[1].add(new pair(0, -2));
        adj[1].add(new pair(2, 1));
        adj[2].add(new pair(1, -1));
        adj[2].add(new pair(3, -2));
        adj[3].add(new pair(2, 2));
        adj[3].add(new pair(0, -1));
        adj[0].add(new pair(3, 1));
        for (int i = 0; i < n; i++) {
            if (flg == 1)
 
                // If for any vertex, flg is true,
                // then the distance is not equal
                break;
 
            if (vis[i] != 1)
 
                // Calling the DFS function if
                // the vertex is not visited
                dfs(i);
        }
        if (flg == 1)
            System.out.print("No" + "\n");
        else
            System.out.print("Yes" + "\n");
    }
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the above approach
adj = [[] for i in range(100005)]
 
# Initialize distance and visited array
vis = [0]*100005
dist = [0]*100005
flg = 0
 
# Function to perform dfs and check
# For a given vertex If the distance
# for all the paths is equal or not
def dfs(curr):
    vis[curr] = 1
    for i in range(len(adj[curr])):
 
        # Checking the next node to reach
        # is visited or not
        if (vis[adj[curr][i][0]]):
 
            # Case 2: comparing the distance
            if (dist[adj[curr][i][0]]
                != dist[curr] + adj[curr][i][1]):
                flg = 1
        else:
 
            # Case 1: Adding the distance
            # and updating the array
            dist[adj[curr][i][0]] = dist[curr]+ adj[curr][i][1]
 
            # Calling the function again with the
            # same node
            dfs(adj[curr][i][0])
 
# Driver code
 
n = 4
m = 4
flg = 0
 
# Creating the graph as mentioned
# in example 1
adj[0].append([1, 2])
adj[1].append([0, -2])
adj[1].append([2, 1])
adj[2].append([1, -1])
adj[2].append([3, -2])
adj[3].append([2, 2])
adj[3].append([0, -1])
adj[0].append([3, 1])
 
for i in range(n):
    if (flg):
 
        # If for any vertex, flg is true,
        # then the distance is not equal
        break
 
    if (vis[i] == 0):
 
        # Calling the DFS function if
        # the vertex is not visited
        dfs(i)
 
if (flg):
    print("No")
else:
    print("Yes")
     
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
using System.Collections.Generic;
 
class GFG {
    class pair {
        public int first, second;
        public pair(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }
  
    static List<pair>[] adj = new List<pair>[100005];
  
    // Initialize distance and visited array
    static int[] vis = new int[100005];
    static int[] dist = new int[100005];
    static int flg;
  
    // Function to perform dfs and check
    // For a given vertex If the distance
    // for all the paths is equal or not
    static void dfs(int curr) {
        vis[curr] = 1;
        for (int i = 0; i < adj[curr].Count; i++) {
  
            // Checking the next node to reach
            // is visited or not
            if (vis[adj[curr][i].first] > 0) {
  
                // Case 2: comparing the distance
                if (dist[adj[curr][i].first] !=
                    dist[curr] + adj[curr][i].second)
                    flg = 1;
            } else {
  
                // Case 1: Adding the distance
                // and updating the array
                dist[adj[curr][i].first] =
                    dist[curr] + adj[curr][i].second;
  
                // Calling the function again with the
                // same node
                dfs(adj[curr][i].first);
            }
        }
    }
  
    // Driver code
    public static void Main(String[] args) {
        int n = 4;
        flg = 0;
        for (int i = 0; i < adj.Length; i++) {
            adj[i] = new List<pair>();
        }
        // Creating the graph as mentioned
        // in example 1
        adj[0].Add(new pair(1, 2));
        adj[1].Add(new pair(0, -2));
        adj[1].Add(new pair(2, 1));
        adj[2].Add(new pair(1, -1));
        adj[2].Add(new pair(3, -2));
        adj[3].Add(new pair(2, 2));
        adj[3].Add(new pair(0, -1));
        adj[0].Add(new pair(3, 1));
        for (int i = 0; i < n; i++) {
            if (flg == 1)
  
                // If for any vertex, flg is true,
                // then the distance is not equal
                break;
  
            if (vis[i] != 1)
  
                // Calling the DFS function if
                // the vertex is not visited
                dfs(i);
        }
        if (flg == 1)
            Console.Write("No" + "\n");
        else
            Console.Write("Yes" + "\n");
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation of the above approach
let  adj = new Array(100005);
 
// Initialize distance and visited array
let vis = new Array(100005);
let dist = new Array(100005);
for(let i = 0; i < 100005; i++)
{
    vis[i] = 0;
    dist[i] = 0;
}
 
let flg;
 
// Function to perform dfs and check
// For a given vertex If the distance
// for all the paths is equal or not
function dfs(curr)
{
    vis[curr] = 1;
    for(let i = 0; i < adj[curr].length; i++)
    {
         
        // Checking the next node to reach
        // is visited or not
        if (vis[adj[curr][i][0]] > 0)
        {
             
            // Case 2: comparing the distance
            if (dist[adj[curr][i][0]] != 
                dist[curr] + adj[curr][i][1])
                flg = 1;
        }
        else
        {
 
            // Case 1: Adding the distance
            // and updating the array
            dist[adj[curr][i][0]] =  dist[curr] +
                                      adj[curr][i][1];
 
            // Calling the function again with the
            // same node
            dfs(adj[curr][i][0]);
        }
    }
}
 
// Driver code
let n = 4, m = 4;
flg = 0;
for(let i = 0; i < adj.length; i++)
{
    adj[i] = [];
}
 
// Creating the graph as mentioned
// in example 1
adj[0].push([1, 2]);
adj[1].push([0, -2]);
adj[1].push([2, 1]);
adj[2].push([1, -1]);
adj[2].push([3, -2]);
adj[3].push([2, 2]);
adj[3].push([0, -1]);
adj[0].push([3, 1]);
for(let i = 0; i < n; i++)
{
    if (flg == 1)
 
        // If for any vertex, flg is true,
        // then the distance is not equal
        break;
 
    if (vis[i] != 1)
 
        // Calling the DFS function if
        // the vertex is not visited
        dfs(i);
}
 
if (flg == 1)
    document.write("No" + "<br>");
else
    document.write("Yes" + "<br>");
 
// This code is contributed by patel2127
 
</script>
Output: 
Yes

 

Time Complexity: O(V + E) where V is no of vertices and E is no of edges
 

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