《数据结构课程设计》图结构练习：List Component

7-1 List Components (30 分)

 

For a given undirected graph with N vertices and E edges, please list all the connected components by both DFS (Depth First Search) and BFS (Breadth First Search). Assume that all the vertices are numbered from 0 to N-1. While searching, assume that we always start from the vertex with the smallest index, and visit its adjacent vertices in ascending order of their indices.

Input Specification:

Each input file contains one test case. For each case, the first line gives two integers N (0<N≤10) and E, which are the number of vertices and the number of edges, respectively. Then E lines follow, each described an edge by giving the two ends. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in each line a connected component in the format { v​1​​ v​2​​ ... v​k​​ }. First print the result obtained by DFS, then by BFS.

Sample Input:

8 6
0 7
0 1
2 0
4 1
2 4
3 5

Sample Output:

{ 0 1 4 2 7 }
{ 3 5 }
{ 6 }
{ 0 1 2 7 4 }
{ 3 5 }
{ 6 }

题目分析：考察图的深度优先遍历、广度优先遍历。由于题中给出了结点数N是小于等于10的，所以采用领接矩阵的方法来做。下面对输入输出样例做简要解释：

输入：结点数（N=8），边数（E=6）。接下来6行分别显示每条边的信息，即两个端点上的数字。注意这里是无向图。
    
输出：首先显示深度优先遍历的结果（DFS），然后显示广度优先遍历的结果（BFS）。注意，此处深度和广度优先遍历都出现了多行的情况。主要是因为图中不止一个连通分量。
    此外，一个结点同时连多个结点，按照结点data的大小升序输出。

代码如下：

#include<iostream>
#include<vector>
#include<queue>

#define MAXSIZE 10
using namespace std;

//深度优先遍历
void DFS(int (*data)[MAXSIZE], int i, bool *visited)  
{
    if (visited[i] == true)
    {
        return;
    }
    cout << i << " ";
    visited[i] = true;

    for (int j = 0; j < MAXSIZE; j++)
    {
        if (data[i][j] == 1 && visited[j] == false)
        {
            DFS(data, j, visited);
        }
    }
}

//广度优先遍历
void BFS(int (*data)[MAXSIZE], int i, bool *visited) 
{
    if (visited[i] == true)
    {
        return;
    }

    queue<int> myQueue;
    myQueue.push(i);
    visited[i] = true;

    while (myQueue.size() != 0)
    {
        int x = myQueue.front();
        cout << x << " ";
        myQueue.pop();
        for (int i = 0; i < MAXSIZE; i++)
        {
            if (data[x][i] == 1 && visited[i] == false)
            {
                visited[i] = true;
                myQueue.push(i);
            }
        }
    }
}

int main()
{
    int data[MAXSIZE][MAXSIZE] = { 0 };
    bool visitedDFS[MAXSIZE] = { false };
    bool visitedBFS[MAXSIZE] = { false };

    int N, E;
    cin >> N >> E;

    //用领接矩阵存储图
    for (int i = 0; i < E; i++)
    {
        int a, b;
        cin >> a >> b;
        data[a][b] = 1;
        data[b][a] = 1;
    }

    //深度优先遍历，从数字最小的结点开始遍历
    for (int i = 0; i < N; i++)
    {
        if (!visitedDFS[i])
        {
            cout << "{"<<" ";
            DFS(data, i, visitedDFS);  //DFS定义中(*data),否则data处标记错误
            cout << "}" << endl;
        }
    }

    //广度优先遍历，从数字最小的结点开始遍历
    for (int i = 0; i < N; i++)
    {
        if (!visitedBFS[i])
        {
            cout << "{"<<" ";
            BFS(data, i, visitedBFS);  //BFS定义中(*data),否则data处标记错误
            cout << "}" << endl;
        }
    }

    return 0;
}

声明：仅作为个人学习记录，对其他代码有所参考，不作其他用途。