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
..................
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
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