forked from Algo-Phantoms/Algo-Tree
-
Notifications
You must be signed in to change notification settings - Fork 0
/
BFS.cpp
124 lines (93 loc) · 2.04 KB
/
BFS.cpp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
/*
The BFS algorithm works similar to level – order traversal of the trees. Like level – order
traversal, BFS also uses queues. In fact, level – order traversal got inspired from BFS. BFS
works level by level. Initially, BFS starts at a given vertex, which is at level 0. In the first stage it
visits all vertices at level 1 (that means, vertices whose distance is 1 from the start vertex of the
graph). In the second stage, it visits all vertices at the second level. These new vertices are the
ones which are adjacent to level 1 vertices.
BFS continues this process until all the levels of the graph are completed. Generally queue data
structure is used for storing the vertices of a level.
*/
#include<iostream>
#include<map>
#include<string>
#include<queue>
#include<list>
using namespace std;
// for generic
template<typename T>
class Graph{
map<T, list<T> > m;
public:
void AddEdge(T src, T dest, bool nondirectional = true){
m[src].push_back(dest);
// bi-directional
if(nondirectional){
m[dest].push_back(src);
}
}
void print(){
for(auto i : m){
cout << i.first<<" -> ";
for(auto j: i.second){
cout<<j<<" -> ";
}
cout<<"\n";
}
}
// like level order traversal
void BFS(T src){
queue<T> q;
q.push(src);
map<T, bool> visited;
visited[src] = true;
while(!q.empty()){
T temp = q.front();
cout<<temp<<" ";
q.pop();
// traverse list of neighbours
for(auto neigh : m[temp]){
// if not visited
if(!visited[neigh]){
q.push(neigh);
visited[neigh] = true;
}
}
}
}
};
int main(){
Graph<int> g;
//no. of edges
int n;
cin >> n;
int u, v;
for(int i=0; i<n; i++){
cin >> u >> v;
g.AddEdge(u,v);
}
g.BFS(0);
return 0;
}
/*
Test Case :
Input : 7
0 1
0 4
1 2
2 3
2 4
3 4
3 5
Output : 0 1 4 2 3 5
Input : 6
0 1
0 2
1 2
2 0
2 3
3 3
Output : 0 1 2 3
Time Complexity: O(V+E) where V is number of vertices in the graph and E is number of edges in the graph.
Space complexity : O(V)
*/