BFS using vectors & queue as per the algorithm of CLRS in C Program?


In CLRS book the BFS algorithm is described using vectors and queue. We have to implement that algorithm using C++ STL. Let us see the algorithm at first.

Algorithm

BFS(G, s) −

begin
   for each vertex u in G.V - {s}, do
      u.color := white
      u.d := infinity
      u.p := NIL
   done
   s.color := green
   s.d := 0
   s.p := NIL
   Q := NULL
   insert s into Q
   while Q is not null, do
      u = delete from Q
      for each v in adjacent to u, do
         if v.color = white
            v.color := green
            v.d := u.d + 1
            v.p := u
            insert v into Q
         end if
      done
      u.color = dark_green
   done
end

Example

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
vector<string> colour;
vector<int> dist;
vector<int> par;
void addEdge(vector <int> g[], int u, int v) { //add edge to form the graph
   g[u].push_back(v);
   g[v].push_back(u);
}
void BFS(vector <int> g[], int s) {
   queue<int> q;
   q.push(s); //insert source
   dist[s] = 0;
   colour[s] = "gray";
   while (!q.empty()) {
      int u = q.front(); //top element from queue, then delete it
      q.pop();
      cout << u << " ";
      for (auto i = g[u].begin(); i != g[u].end(); i++) {
         if (colour[*i] == "white") { //white is unvisited node
            colour[*i] = "gray"; //gray is visited but not completed
            dist[*i] = dist[u] + 1;
            par[*i] = u;
            q.push(*i);
         }
      }
      colour[u] = "black"; //black is completed node
   }
}
void BFSAlgo(vector <int> g[], int n) {
   colour.assign(n, "white"); //put as unvisited
   dist.assign(n, 0);
   par.assign(n, -1);
   for (int i = 0; i < n; i++)
      if (colour[i] == "white")
   BFS(g, i);
}
int main() {
   int n = 7;
   vector <int> g[n];
   addEdge(g, 0, 1);
   addEdge(g, 0, 2);
   addEdge(g, 1, 3);
   addEdge(g, 1, 4);
   addEdge(g, 2, 5);
   addEdge(g, 2, 6);
   BFSAlgo(g, n);
}

Output

0 1 2 3 4 5 6

Updated on: 20-Aug-2019

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