Analysis:
1. Depth first is more suitable for situations where the goal is relatively clear, and the main purpose is to find the goal
2. Breadth first is more suitable for situations where the relatively optimal solution is found when the traversal range is continuously expanded
Therefore, BFS-breadth first traversal is selected here
Idea: To find the solution with the least number of transfers here is to find the shortest path from the V0 vertex to the V4 vertex.
1. First list which cities have routes: (whether there are edges between the two vertices)
v0------v1
v0------v2
v1------v2
v1---- --v3
v2------v3
v2------v4
v3------v4
2. During the breadth-first traversal process, when the vertex reached is v4, the breadth-first traversal is exited. The minimum number of times
The user enters four values: there are several cities, there are several routes, starting city and destination city
Return: Minimum number of transfers and transfer plan
01234 is used here to represent v0, v1, v2, v3, v4
#include<iostream>
using namespace std;
#include<queue>
const int MAX = 10; //图的最大顶点个数为10
typedef int DataType;
class Graph
{
private:
//边的个数
int arcNum;
//顶点个数
int vertexNum;
//定义一个存放顶点的一位数组
DataType vertex[MAX];
//定义一个存放顶点间的边关系的二维数组
int arc[MAX][MAX];
//访问数组
int visit[MAX];
public:
//v[]数组存放用户输入的一维数组的顶点数据，n表示顶点个数，e是边的个数
Graph(DataType v[], int n, int e,int VI[],int VJ[]);
//BFS----广度优先遍历
int BFS();
};
//有参构造函数的实现
Graph::Graph(DataType v[], int n, int e,int VI[],int VJ[])
{
//初始化顶点个数
vertexNum = n;
//初始化边的个数
arcNum = e;
//初始化顶点数组
for (int i = 0; i <n; i++)
{
vertex[i] = v[i];
}
//初始化边数组
for (int i = 0; i < MAX; i++)
for (int j = 0; j < MAX; j++)
arc[i][j] = 0;
//初始化访问数组
for (int i = 0; i < MAX; i++)
visit[i] = 0;//一开始所有节点都处于未被访问的状态
//由用户来决定，哪两个顶点之间存在边
int vi=0, vj=0;
for (int i = 0; i < arcNum; i++)
{
//两个顶点之间的边关系
int vi = VI[i];
int vj = VJ[i];
cout << vi << "<---->" << vj << endl;
//这是无向图的边初始化标志
arc[vi][vj] = 1;//有边的标志
arc[vj][vi] = 1;
}
}
//BFS-----广度优先遍历
int Graph::BFS()
{
int num = 0;//记录转机次数
queue<DataType> q;//队列存储的是顶点信息
//外层for循环，检查是否每个节点都被访问过，防止存在节点未被访问过
for (int i = 0; i < vertexNum; i++)
{
if (visit[i]==0)//如果当前节点没有被访问过就进行处理
{
//设置当前节点被访问过
visit[i] == 1;
cout << vertex[i]<<" ";
//说明找到了四号城市，退出当前遍历过程
if (vertex[i] == 3)
{
return num;
}
//将此顶点入队
q.push(vertex[i]);
//若当前队列不为空---表示当前顶点还存在邻接点没有被访问过
while (!q.empty())
{
q.pop();//队头元素出队
//遍历当前顶点在邻接矩阵中当前行，找找是否存在未被访问过的顶点
for (int j = 0; j < vertexNum; j++)
{
//当前两个顶点之间有边 当前顶点的邻接点未被访问过
if (arc[i][j] == 1 && visit[j] == 0)
{
//将找到的此节点标志设置为已经访问
visit[j] = 1;
//输出这两个被边连接的顶点
cout <<vertex[j] <<" ";
//找到一个邻接点，次数++
num++;
//将找到的此节点入队
//每次把当前顶点入队都是为了得到它的邻接点，并判断是否被访问过
q.push(vertex[j]);
}
//找到的当前邻接点为4号城市,退出遍历
if (vertex[j] == 3)
{
return num;
}
}
}
}
}
//0号城市和4号城市之间不存在航线可以抵达
return -1;
}
//测试-------------------
void test()
{
//存储两个顶点之间边关系的数组
int VI[14] = {
0,0,2,2,2,3,3};
int VJ[7] = {
1,2,1,4,3,1,4 };
DataType v[5] = {
0,1,2,3,4 };
cout << "存在航线的城市有：" << endl;
Graph p(v, 5, 7,VI,VJ);
cout << "输出所有城市:" << endl;
int num=p.BFS();
cout << endl;
cout << "0号到3号城市之间的最少转机次数为："<<num << endl;
}
int main()
{
test();
system("pause");
return 0;
}
