BFS traverse the graph - graph theory foundation

1. Analysis

Continued on a blog, with DFS traversal Figure: Graph theory foundation - Traverse map of DFS

We also can be used BFS traversal of this graph, traversal time stamp mark on it:

2. Algorithm Design 

 Using the breadth-first search to traverse this figure is as follows:

First, a vertex not been accessed as a start vertex, such as vertex number 1 as a starting point. One vertex and then put into the queue, then the vertices adjacent to vertex number 1, i.e., 2,3,5 unvisited No. vertex again put into the queue, as shown below:

Then again on the 2nd vertex is not adjacent visited No. 4 vertex added to the queue:

So far, all the vertices are traversed, the end of the BFS.

3. Source Code 

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int N=111;
const int INF=9999999;
int main()
{
    int edge[N][N];
    int marked[N]= {0};
    int n;
    int m;
    int point_a;
    int point_b;
    int cur;
    int que[11111];
    int head;
    int tail;
    cout << "请输入顶点个数n和边的条数m："<< endl;
    cin >> n>> m;
    //初始化
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            if(i==j)
            {
                edge[i][j]=0;
            }
            else
            {
                edge[i][j]=INF;
            }
        }
    }
    cout << "请输入连接边的两个顶点：" << endl;
    for(int i=1; i<=m; i++)
    {
        cin >> point_a >> point_b;
        edge[point_a][point_b]=1;
        edge[point_b][point_a]=1;
    }
    //队列初始化
    head=1;
    tail=1;
    que[tail]=1;
    tail++;
    marked[1]=1;
    while(head<tail && tail<=n)
    {
        cur=que[head];//当前访问的结点
        for(int i=1; i<=n; i++)
        {
            if(edge[cur][i]==1 && marked[i]==0)
            {
                que[tail]=i;
                tail++;
                marked[i]=1;
            }
            if(tail > n)
            {
                break;
            }
        }
        head++;//出列
    }
    cout << "遍历结点的顺序是："<< endl;
    for(int i=1; i<tail; i++)
    {
        cout <<que[i];
    }
    cout << endl;
    return 0;
}

 

4. Test results