技術標籤：演算法作業c++圖論圖解法

分別用廣度優先搜尋得到以每個頂點為根節點的最小半徑的生成樹，然後從這些生成樹中找出半徑最小的作為本題的解。

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

using namespace std;

struct SpanningTree		//生成樹的半徑及其層次遍歷序列
{
	int radius = 0;		//半徑
	vector <int> hierarchically_traverse_seq;	//層次遍歷序列
};

int** build_adjacency_matrix(int n)		//輸入一個圖，用鄰接矩陣表示，以輸入兩個負數結束
{
	int** adjacency_matrix = new int* [n];		//初始化鄰接矩陣，使其每個元素都為0
	for (int i = 0; i < n; i++)
	{
		adjacency_matrix[i] = new int[n];
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			adjacency_matrix[i][j] = 0;
		}
	}
	int a, b;				//輸入邊，a和b分別是邊所連線的兩個頂點
	cout << "Please input each edge" << endl;
    cin >> a >> b;
	while (a >= 0 && b >= 0)
	{
		adjacency_matrix[a][b] = adjacency_matrix[b][a] = 1;
		cin >> a >> b;
	}
	return adjacency_matrix;
}

/*以序號為initial_vertex的頂點作為根節點，通過廣度優先搜尋建造半徑最小的生成樹*/
SpanningTree hierarchically_traverse(int** adjacency_matrix, int initial_vertex, int n)
{
	SpanningTree st;
	queue <int> vertex_queue;		//頂點佇列
	vector <int> dist(n, 0);		//各個頂點到根節點的距離
	vector <int> visited(n, 0);		//記錄每個頂點是否被訪問過，1表示訪問過，0表示沒訪問過

	/*廣度優先搜尋*/
	st.hierarchically_traverse_seq.push_back(initial_vertex);
	vertex_queue.push(initial_vertex);
	visited[initial_vertex] = 1;
	while (!vertex_queue.empty())
	{
		int head = vertex_queue.front();
		for (int j = 0; j < n; j++)
		{
			if (adjacency_matrix[head][j] == 1 && visited[j] == 0)
			{
				vertex_queue.push(j);
				st.hierarchically_traverse_seq.push_back(j);
				visited[j] = 1;
				dist[j] = dist[head] + 1;	//計算各個頂點到根節點的距離
			}
		}
		vertex_queue.pop();
	}

	/*計算生成樹的半徑*/
	st.radius = dist[0];
	for (int i = 1; i < dist.size(); i++)
	{
		if (st.radius < dist[i])
			st.radius = dist[i];
	}
	return st;
}

int main(void)
{
	int n;      //頂點數
    cout << "Please input the quantity of vertexes: ";
	cin >> n;
	int** adjacency_matrix = build_adjacency_matrix(n);  //建立鄰接矩陣
	vector <SpanningTree> st(n);	//用於分別記錄以每個頂點為根節點的生成樹
	for (int i = 0; i < n; i++)
		st[i] = hierarchically_traverse(adjacency_matrix, i, n);
	
	/*找出半徑最小的生成樹*/
	int min_radius = 0;
	for (int i = 1; i < n; i++)
	{
		if (st[i].radius < st[min_radius].radius)
			min_radius = i;
	}
	cout << endl;
	cout << "The minimal radius is " << st[min_radius].radius << endl;
	cout << "The hierarchically traverse of the spanning tree is: ";
	for (int i = 0; i < n; i++)
		cout << st[min_radius].hierarchically_traverse_seq[i] << ' ';
	cout << endl;

	return 0;
}