0-1 BFS (Shortest Path in a Binary Weight Graph)

Given a graph where every edge has weight as either 0 or 1. A source vertex is also given in the graph. Find the shortest path from source vertex to every other vertex.
Example:

Input : Source Vertex = 0 and below graph 


Output : Shortest distances from given source
         0 0 1 1 2 1 2 1 2

Explanation : 
Shortest distance from 0 to 0 is 0
Shortest distance from 0 to 1 is 0
Shortest distance from 0 to 2 is 1
..................

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

In normal BFS of a graph all edges have equal weight but in 0-1 BFS some edges may have 0 weight and some may have 1 weight. In this we will not use bool array to mark visited nodes but at each step we will check for the optimal distance condition. We use double ended queue to store the node. While performing BFS if a edge having weight = 0 is found node is pushed at front of double ended queue and if a edge having weight = 1 is found, it is pushed at back of double ended queue.

The approach is similar to Dijkstra that the if the shortest distance to node is relaxed by the previous node then only it will be pushed in the queue.

The above idea works in all cases, when pop a vertex (like Dijkstra), it is the minimum weight vertex among remaining vertices. If there is a 0 weight vertex adjacent to it, then this adjacent has same distance. If there is a 1 weight adjacent, then this adjacent has maximum distance among all vertices in dequeue (because all other vertices are either adjacent of currently popped vertex or adjacent of previously popped vertices).

Below is the implementation of the above idea.

C++

// C++ program to implement single source 
// shortest path for a Binary Graph 
#include<bits/stdc++.h> 
using namespace std; 
  
/* no.of vertices */
#define V 9 
  
// a structure to represent edges 
struct node 
{ 
    // two variable one denote the node 
    // and other the weight 
    int to, weight; 
}; 
  
// vector to store edges 
vector <node> edges[V]; 
  
// Prints shortest distance from given source to 
// every other vertex 
void zeroOneBFS(int src) 
{ 
    // Initialize distances from given source 
    int dist[V]; 
    for (int i=0; i<V; i++) 
        dist[i] = INT_MAX; 
  
    // double ende queue to do BFS. 
    deque <int> Q; 
    dist[src] = 0; 
    Q.push_back(src); 
  
    while (!Q.empty()) 
    { 
        int v = Q.front(); 
        Q.pop_front(); 
  
        for (int i=0; i<edges[v].size(); i++) 
        { 
            // checking for the optimal distance 
            if (dist[edges[v][i].to] > dist[v] + edges[v][i].weight) 
            { 
                dist[edges[v][i].to] = dist[v] + edges[v][i].weight; 
  
                // Put 0 weight edges to front and 1 weight 
                // edges to back so that vertices are processed 
                // in increasing order of weights. 
                if (edges[v][i].weight == 0) 
                    Q.push_front(edges[v][i].to); 
                else
                    Q.push_back(edges[v][i].to); 
            } 
        } 
    } 
  
    // printing the shortest distances 
    for (int i=0; i<V; i++) 
        cout << dist[i] << " "; 
} 
  
void addEdge(int u, int v, int wt) 
{ 
   edges[u].push_back({v, wt}); 
   edges[v].push_back({u, wt}); 
} 
  
// Driver function 
int main() 
{ 
    addEdge(0, 1, 0); 
    addEdge(0, 7, 1); 
    addEdge(1, 7, 1); 
    addEdge(1, 2, 1); 
    addEdge(2, 3, 0); 
    addEdge(2, 5, 0); 
    addEdge(2, 8, 1); 
    addEdge(3, 4, 1); 
    addEdge(3, 5, 1); 
    addEdge(4, 5, 1); 
    addEdge(5, 6, 1); 
    addEdge(6, 7, 1); 
    addEdge(7, 8, 1); 
    int src = 0;//source node 
    zeroOneBFS(src); 
    return 0; 
} 

Java

// Java Program to implement 0-1 BFS 
import java.util.ArrayDeque; 
import java.util.ArrayList; 
import java.util.Deque; 
  
public class ZeroOneBFS { 
    private static class Node { 
        int to; // the ending vertex 
        int weight; // the weight of the edge 
          
        public Node(int to, int wt) { 
            this.to = to; 
            this.weight = wt; 
        } 
    } 
      
    private static final int numVertex = 9; 
    private ArrayList<Node>[] edges = new ArrayList[numVertex]; 
      
    public ZeroOneBFS() { 
        for (int i = 0; i < edges.length; i++) { 
            edges[i] = new ArrayList<Node>(); 
        } 
    } 
      
    public void addEdge(int u, int v, int wt) { 
        edges[u].add(edges[u].size(), new Node(v, wt)); 
        edges[v].add(edges[v].size(), new Node(u, wt)); 
    } 
      
    public void zeroOneBFS(int src) { 
  
        // initialize distances from given source 
        int[] dist = new int[numVertex]; 
        for (int i = 0; i < numVertex; i++) { 
            dist[i] = Integer.MAX_VALUE; 
        } 
          
        // double ended queue to do BFS 
        Deque<Integer> queue = new ArrayDeque<Integer>(); 
        dist[src] = 0; 
        queue.addLast(src); 
          
        while (!queue.isEmpty()) { 
            int v = queue.removeFirst(); 
            for (int i = 0; i < edges[v].size(); i++) { 
  
                // checking for optimal distance 
                if (dist[edges[v].get(i).to] >  
                        dist[v] + edges[v].get(i).weight) { 
  
                    // update the distance 
                    dist[edges[v].get(i).to] = 
                          dist[v] + edges[v].get(i).weight; 
  
                    // put 0 weight edges to front and 1 
                    // weight edges to back so that vertices 
                    // are processed in increasing order of weight 
                    if (edges[v].get(i).weight == 0) { 
                        queue.addFirst(edges[v].get(i).to); 
                    } else { 
                        queue.addLast(edges[v].get(i).to); 
                    } 
                } 
            } 
        } 
          
        for (int i = 0; i < dist.length; i++) { 
            System.out.print(dist[i] + " "); 
        } 
    } 
      
    public static void main(String[] args) { 
        ZeroOneBFS graph = new ZeroOneBFS(); 
        graph.addEdge(0, 1, 0); 
        graph.addEdge(0, 7, 1); 
        graph.addEdge(1, 7, 1); 
        graph.addEdge(1, 2, 1); 
        graph.addEdge(2, 3, 0); 
        graph.addEdge(2, 5, 0); 
        graph.addEdge(2, 8, 1); 
        graph.addEdge(3, 4, 1); 
        graph.addEdge(3, 5, 1); 
        graph.addEdge(4, 5, 1); 
        graph.addEdge(5, 6, 1); 
        graph.addEdge(6, 7, 1); 
        graph.addEdge(7, 8, 1); 
        int src = 0;//source node 
        graph.zeroOneBFS(src); 
        return; 
    } 
} 

Python3

# Python3 program to implement single source 
# shortest path for a Binary Graph 
from sys import maxsize as INT_MAX 
from collections import deque 
  
# no.of vertices 
V = 9
  
# a structure to represent edges 
class node: 
    def __init__(self, to, weight): 
  
        # two variable one denote the node 
        # and other the weight 
        self.to = to 
        self.weight = weight 
  
# vector to store edges 
edges = [0] * V 
for i in range(V): 
    edges[i] = [] 
  
# Prints shortest distance from  
# given source to every other vertex 
def zeroOneBFS(src: int): 
  
    # Initialize distances from given source 
    dist = [0] * V 
    for i in range(V): 
        dist[i] = INT_MAX 
  
    # double ende queue to do BFS. 
    Q = deque() 
    dist[src] = 0
    Q.append(src) 
  
    while Q: 
        v = Q[0] 
        Q.popleft() 
  
        for i in range(len(edges[v])): 
  
            # checking for the optimal distance 
            if (dist[edges[v][i].to] >  
                dist[v] + edges[v][i].weight): 
                dist[edges[v][i].to] = dist[v] + edges[v][i].weight 
  
                # Put 0 weight edges to front and 1 weight 
                # edges to back so that vertices are processed 
                # in increasing order of weights. 
                if edges[v][i].weight == 0: 
                    Q.appendleft(edges[v][i].to) 
                else: 
                    Q.append(edges[v][i].to) 
  
    # printing the shortest distances 
    for i in range(V): 
        print(dist[i], end = " ") 
    print() 
  
def addEdge(u: int, v: int, wt: int): 
    edges[u].append(node(v, wt)) 
    edges[u].append(node(v, wt)) 
  
# Driver Code 
if __name__ == "__main__": 
  
    addEdge(0, 1, 0) 
    addEdge(0, 7, 1) 
    addEdge(1, 7, 1) 
    addEdge(1, 2, 1) 
    addEdge(2, 3, 0) 
    addEdge(2, 5, 0) 
    addEdge(2, 8, 1) 
    addEdge(3, 4, 1) 
    addEdge(3, 5, 1) 
    addEdge(4, 5, 1) 
    addEdge(5, 6, 1) 
    addEdge(6, 7, 1) 
    addEdge(7, 8, 1) 
  
    # source node 
    src = 0
    zeroOneBFS(src) 
  
# This code is contributed by 
# sanjeev2552 

Output:

0 0 1 1 2 1 2 1 2

This problem can also be solved by Dijkstra but the time complexity will be O(E + V Log V) whereas by BFS it will be O(V+E).

Reference :
http://codeforces.com/blog/entry/22276

This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

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