Implementing Water Supply Problem using Breadth First Search
Given N cities that are connected using N-1 roads. Between Cities [i, i+1], there exists an edge for all i from 1 to N-1.
The task is to set up a connection for water supply. Set the water supply in one city and water gets transported from it to other cities using road transport. Certain cities are blocked which means that water cannot pass through that particular city. Determine the maximum number of cities to which water can be supplied.
Input format:
- The first line contains an integer >strong>N denoting the number of cities.
- The next N-1 lines contain two space-separated integers u v denoting a road between
city u and v. - The next line contains N space-separated integers where it is 1 if the ith city is
blocked, else it is 0.
Examples:
Input :
4
1 2
2 3
3 4
0 1 1 0
Output :
2
Explanation : If city 1 is chosen, then water is supplied from
city 1 to 2. If city 4 is chosen, water is supplied from city 4 to 3
hence maximum of 2 cities can be supplied with water.
Input :
7
1 2
2 3
3 4
4 5
5 6
6 7
0 1 1 0 0 0 0
Output :
5
Explanation : If city 1 is chosen than water is supplied from
city 1 to 2 or if city 4 is chosen water is supplied from city 4 to
3, 5, 6 and 7 hence maximum of 5 cities are supplied with water.
Approach:
In this post, a BFS based solution is discussed.
We run a breadth-first search on each city and check for two things: The city is not blocked and the city is not visited. If both these conditions return true then we run a breadth-first search from that city and count the number of cities up to which water can be supplied.
This solution can also be achieved using a depth-first search.
Below is the implementation of the above approach:
C++
// C++ program to solve water // supply problem using BFS #include <iostream> #include <vector> #include <queue> using namespace std; // Function to perform BFS int bfsUtil( int v[], bool vis[], vector< int > adj[], int src) { // Mark current source visited vis[src] = true ; queue< int > q; //Queue for BFS q.push(src); // Push src to queue int count = 0; while (!q.empty()) { int p = q.front(); for ( int i = 0; i < adj[p].size(); i++) { // When the adjacent city not visited and // not blocked, push city in the queue. if (!vis[adj[p][i]] && v[adj[p][i]] == 0) { count++; vis[adj[p][i]] = true ; q.push(adj[p][i]); } // when the adjacent city is not visited // but blocked so the blocked city is // not pushed in queue else if (!vis[adj[p][i]] && v[adj[p][i]] == 1) { count++; } } q.pop(); } return count + 1; } // Utility function to perform BFS int bfs( int N, int v[], vector< int > adj[]) { bool vis[N + 1]; int max = 1, res; // marking visited array false for ( int i = 1; i <= N; i++) vis[i] = false ; // Check for each and every city for ( int i = 1; i <= N; i++) { // Checks that city is not blocked // and not visited. if (v[i] == 0 && !vis[i]) { res = bfsUtil(v, vis, adj, i); if (res > max) { max = res; } } } return max; } // Driver Code int main() { int N = 4; // Denotes the number of cities vector< int > adj[N + 1]; int v[N + 1]; // Adjacency list denoting road // between city u and v adj[1].push_back(2); adj[2].push_back(1); adj[2].push_back(3); adj[3].push_back(2); adj[3].push_back(4); adj[4].push_back(3); // array for storing whether ith // city is blocked or not v[1] = 0; v[2] = 1; v[3] = 1; v[4] = 0; cout<<bfs(N, v, adj); return 0; } |
Java
// Java program to solve water // supply problem using BFS import java.util.*; class GFG{ // Function to perform BFS static int bfsUtil( int v[], boolean vis[], Vector<Integer> adj[], int src) { // Mark current source visited vis[src] = true ; // Queue for BFS Queue<Integer> q = new LinkedList<>(); // Push src to queue q.add(src); int count = 0 ; while (!q.isEmpty()) { int p = q.peek(); for ( int i = 0 ; i < adj[p].size(); i++) { // When the adjacent city not // visited and not blocked, push // city in the queue. if (!vis[adj[p].get(i)] && v[adj[p].get(i)] == 0 ) { count++; vis[adj[p].get(i)] = true ; q.add(adj[p].get(i)); } // When the adjacent city is not visited // but blocked so the blocked city is // not pushed in queue else if (!vis[adj[p].get(i)] && v[adj[p].get(i)] == 1 ) { count++; } } q.remove(); } return count + 1 ; } // Utility function to perform BFS static int bfs( int N, int v[], Vector<Integer> adj[]) { boolean []vis = new boolean [N + 1 ]; int max = 1 , res; // Marking visited array false for ( int i = 1 ; i <= N; i++) vis[i] = false ; // Check for each and every city for ( int i = 1 ; i <= N; i++) { // Checks that city is not blocked // and not visited. if (v[i] == 0 && !vis[i]) { res = bfsUtil(v, vis, adj, i); if (res > max) { max = res; } } } return max; } // Driver Code public static void main(String[] args) { // Denotes the number of cities int N = 4 ; @SuppressWarnings ( "unchecked" ) Vector<Integer> []adj = new Vector[N + 1 ]; for ( int i = 0 ; i < adj.length; i++) adj[i] = new Vector<Integer>(); int []v = new int [N + 1 ]; // Adjacency list denoting road // between city u and v adj[ 1 ].add( 2 ); adj[ 2 ].add( 1 ); adj[ 2 ].add( 3 ); adj[ 3 ].add( 2 ); adj[ 3 ].add( 4 ); adj[ 4 ].add( 3 ); // Array for storing whether ith // city is blocked or not v[ 1 ] = 0 ; v[ 2 ] = 1 ; v[ 3 ] = 1 ; v[ 4 ] = 0 ; System.out.print(bfs(N, v, adj)); } } // This code is contributed by Princi Singh |
Python3
# Python3 program to solve water # supply problem using BFS # Function to perform BFS def bfsUtil(v, vis, adj, src): # Mark current source visited vis[src] = True # Queue for BFS q = [] # Push src to queue q.append(src) count = 0 while ( len (q) ! = 0 ): p = q[ 0 ] for i in range ( len (adj[p])): # When the adjacent city not visited and # not blocked, push city in the queue. if (vis[adj[p][i]] = = False and v[adj[p][i]] = = 0 ): count + = 1 vis[adj[p][i]] = True q.push(adj[p][i]) # when the adjacent city is not visited # but blocked so the blocked city is # not pushed in queue elif (vis[adj[p][i]] = = False and v[adj[p][i]] = = 1 ): count + = 1 q.remove(q[ 0 ]) return count + 1 # Utility function to perform BFS def bfs(N, v, adj): vis = [ 0 for i in range (N + 1 )] mx = 1 # marking visited array false for i in range ( 1 , N + 1 , 1 ): vis[i] = False # Check for each and every city for i in range ( 1 , N + 1 , 1 ): # Checks that city is not blocked # and not visited. if (v[i] = = 0 and vis[i] = = False ): res = bfsUtil(v, vis, adj, i) if (res > mx): mx = res return mx # Driver Code if __name__ = = '__main__' : N = 4 # Denotes the number of cities adj = [[] for i in range (N + 1 )] v = [ 0 for i in range (N + 1 )] # Adjacency list denoting road # between city u and v adj[ 1 ].append( 2 ) adj[ 2 ].append( 1 ) adj[ 2 ].append( 3 ) adj[ 3 ].append( 2 ) adj[ 3 ].append( 4 ) adj[ 4 ].append( 3 ) # array for storing whether ith # city is blocked or not v[ 1 ] = 0 v[ 2 ] = 1 v[ 3 ] = 1 v[ 4 ] = 0 print (bfs(N, v, adj)) # This code is contributed by Bhupendra_Singh |
C#
// C# program to solve water // supply problem using BFS using System; using System.Collections.Generic; class GFG{ // Function to perform BFS static int bfsUtil( int []v, bool []vis, List< int > []adj, int src) { // Mark current source visited vis[src] = true ; // Queue for BFS Queue< int > q = new Queue< int >(); // Push src to queue q.Enqueue(src); int count = 0; while (q.Count != 0) { int p = q.Peek(); for ( int i = 0; i < adj[p].Count; i++) { // When the adjacent city not // visited and not blocked, push // city in the queue. if (!vis[adj[p][i]] && v[adj[p][i]] == 0) { count++; vis[adj[p][i]] = true ; q.Enqueue(adj[p][i]); } // When the adjacent city is not visited // but blocked so the blocked city is // not pushed in queue else if (!vis[adj[p][i]] && v[adj[p][i]] == 1) { count++; } } q.Dequeue(); } return count + 1; } // Utility function to perform BFS static int bfs( int N, int []v, List< int > []adj) { bool []vis = new bool [N + 1]; int max = 1, res; // Marking visited array false for ( int i = 1; i <= N; i++) vis[i] = false ; // Check for each and every city for ( int i = 1; i <= N; i++) { // Checks that city is not blocked // and not visited. if (v[i] == 0 && !vis[i]) { res = bfsUtil(v, vis, adj, i); if (res > max) { max = res; } } } return max; } // Driver Code public static void Main(String[] args) { // Denotes the number of cities int N = 4; List< int > []adj = new List< int >[N + 1]; for ( int i = 0; i < adj.Length; i++) adj[i] = new List< int >(); int []v = new int [N + 1]; // Adjacency list denoting road // between city u and v adj[1].Add(2); adj[2].Add(1); adj[2].Add(3); adj[3].Add(2); adj[3].Add(4); adj[4].Add(3); // Array for storing whether ith // city is blocked or not v[1] = 0; v[2] = 1; v[3] = 1; v[4] = 0; Console.Write(bfs(N, v, adj)); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript program to solve water // supply problem using BFS // Function to perform BFS function bfsUtil(v, vis, adj, src) { // Mark current source visited vis[src] = true ; // Queue for BFS let q = []; // Push src to queue q.push(src); let count = 0; while (q.length > 0) { let p = q[0]; for (let i = 0; i < adj[p].length; i++) { // When the adjacent city not // visited and not blocked, push // city in the queue. if (!vis[adj[p][i]] && v[adj[p][i]] == 0) { count++; vis[adj[p][i]] = true ; q.add(adj[p][i]); } // When the adjacent city is not visited // but blocked so the blocked city is // not pushed in queue else if (!vis[adj[p][i]] && v[adj[p][i]] == 1) { count++; } } q.shift(); } return count + 1; } // Utility function to perform BFS function bfs(N, v, adj) { let vis = new Array(N + 1); let max = 1, res; // Marking visited array false for (let i = 1; i <= N; i++) vis[i] = false ; // Check for each and every city for (let i = 1; i <= N; i++) { // Checks that city is not blocked // and not visited. if (v[i] == 0 && !vis[i]) { res = bfsUtil(v, vis, adj, i); if (res > max) { max = res; } } } return max; } // Denotes the number of cities let N = 4; let adj = new Array(N + 1); for (let i = 0; i < adj.length; i++) adj[i] = []; let v = new Array(N + 1); // Adjacency list denoting road // between city u and v adj[1].push(2); adj[2].push(1); adj[2].push(3); adj[3].push(2); adj[3].push(4); adj[4].push(3); // Array for storing whether ith // city is blocked or not v[1] = 0; v[2] = 1; v[3] = 1; v[4] = 0; document.write(bfs(N, v, adj)); // This code is contributed by divyeshrabadiya07. </script> |
Output:
2