Two coloring Breadth-First Search - graph-algorithm

Question

In a standard BFS implementation, a node can be one of three colors to represent if it is undiscovered, discovered but incomplete, or discovered and completed. Is there a way to implement BFS using only two colors instead of three?

Answer 1

Yes, you can represent it with only two colors. Actually , in probably 99% of problems, you don't need a third color. You need to have an answer to: Is the node X in queue or not?
To answer that question we need to have an array. Let's say we call that array visited.
Values that this array can have are, 0 or 1.
visited[X] = 1, if the node X is in queue(node X is waiting to be processed) or the node is was in queue(which means node X is currently being processed, or was processed and we are done with that node)
visited[X] = 0, if the node X was not yet in queue
Here is a code:
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <queue>
using namespace std;
const int N = 10003;
char visited[N];
queue<int> q;
vector<int> ls[N];
int main() {
int n, m; scanf("%d%d", &n, &m); // number of vertices and number of edges
for(int i = 0; i < m; ++i) {
int x, y; scanf("%d%d", &x, &y);
ls[x].push_back(y);
ls[y].push_back(x);
}
int num_of_components = 0;
for(int i = 1; i <= n; ++i) // iterating through all nodes
if(!visited[i]) { // if we didn't visit node i , then this is a new component
visited[i] = '1'; // mark node as visited
q.push(i); // push node to queue
++num_of_components; // new component was found, so add one
while(!q.empty()) {
int x = q.front();
q.pop();
int sz = ls[x].size();
for(int j = 0; j < sz; ++j) {
int y = ls[x][j];
if(!visited[y]) {
visited[y] = '1';
q.push(y);
}
}
}
}
printf("%d\n", num_of_components);
return 0;
}