最小路径问题 | Dijkstra算法详解(附代码)

来源:AI蜗牛车 本文共3400字,建议阅读6分钟 本文对Dijkstra算法做了一个详细的介绍。 
一、最短路径问题介绍
1、从图中的某个顶点出发到达另外一个顶点的所经过的边的权重和最小的一条路径,称为最短路径。

迪杰斯特拉算法(Dijkstra算法) 弗洛伊德算法(Floyd算法) SPFA算法 
二、Dijkstra算介绍
算法特点
算法的思路
三、Dijkstra算法示例演示







四、Dijkstra算法的代码实现(c++)
Dijkstra.h文件的代码 
/************************************************************//* 程序作者:Willam *//************************************************************///@尽量写出完美的程序using namespace std;/*本程序是使用Dijkstra算法实现求解最短路径的问题采用的邻接矩阵来存储图*///记录起点到每个顶点的最短路径的信息struct Dis {string path;int value;bool visit;Dis() {visit = false;value = 0;path = "";}};class Graph_DG {private:int vexnum; //图的顶点个数int edge; //图的边数int **arc; //邻接矩阵Dis * dis; //记录各个顶点最短路径的信息public://构造函数Graph_DG(int vexnum, int edge);//析构函数~Graph_DG();// 判断我们每次输入的的边的信息是否合法//顶点从1开始编号bool check_edge_value(int start, int end, int weight);//创建图void createGraph();//打印邻接矩阵void print();//求最短路径void Dijkstra(int begin);//打印最短路径void print_path(int);};
Dijkstra.cpp文件的代码 
//构造函数Graph_DG::Graph_DG(int vexnum, int edge) {//初始化顶点数和边数this->vexnum = vexnum;this->edge = edge;//为邻接矩阵开辟空间和赋初值arc = new int*[this->vexnum];dis = new Dis[this->vexnum];for (int i = 0; i < this->vexnum; i++) {arc[i] = new int[this->vexnum];for (int k = 0; k < this->vexnum; k++) {//邻接矩阵初始化为无穷大arc[i][k] = INT_MAX;}}}//析构函数Graph_DG::~Graph_DG() {delete[] dis;for (int i = 0; i < this->vexnum; i++) {delete this->arc[i];}delete arc;}// 判断我们每次输入的的边的信息是否合法//顶点从1开始编号bool Graph_DG::check_edge_value(int start, int end, int weight) {if (start<1 || end<1 || start>vexnum || end>vexnum || weight < 0) {return false;}return true;}void Graph_DG::createGraph() {cout << "请输入每条边的起点和终点(顶点编号从1开始)以及其权重" << endl;int start;int end;int weight;int count = 0;while (count != this->edge) {cin >> start >> end >> weight;//首先判断边的信息是否合法while (!this->check_edge_value(start, end, weight)) {cout << "输入的边的信息不合法,请重新输入" << endl;cin >> start >> end >> weight;}//对邻接矩阵对应上的点赋值arc[start - 1][end - 1] = weight;//无向图添加上这行代码//arc[end - 1][start - 1] = weight;++count;}}void Graph_DG::print() {cout << "图的邻接矩阵为:" << endl;int count_row = 0; //打印行的标签int count_col = 0; //打印列的标签//开始打印while (count_row != this->vexnum) {count_col = 0;while (count_col != this->vexnum) {if (arc[count_row][count_col] == INT_MAX)cout << "∞" << " ";elsecout << arc[count_row][count_col] << " ";++count_col;}cout << endl;++count_row;}}void Graph_DG::Dijkstra(int begin){//首先初始化我们的dis数组int i;for (i = 0; i < this->vexnum; i++) {//设置当前的路径dis[i].path = "v" + to_string(begin) + "-->v" + to_string(i + 1);dis[i].value = arc[begin - 1][i];}//设置起点的到起点的路径为0dis[begin - 1].value = 0;dis[begin - 1].visit = true;int count = 1;//计算剩余的顶点的最短路径(剩余this->vexnum-1个顶点)while (count != this->vexnum) {//temp用于保存当前dis数组中最小的那个下标//min记录的当前的最小值int temp=0;int min = INT_MAX;for (i = 0; i < this->vexnum; i++) {if (!dis[i].visit && dis[i].valuemin = dis[i].value;temp = i;}}//cout << temp + 1 << " "<//把temp对应的顶点加入到已经找到的最短路径的集合中dis[temp].visit = true;++count;for (i = 0; i < this->vexnum; i++) {//注意这里的条件arc[temp][i]!=INT_MAX必须加,不然会出现溢出,从而造成程序异常if (!dis[i].visit && arc[temp][i]!=INT_MAX && (dis[temp].value + arc[temp][i]) < dis[i].value) {//如果新得到的边可以影响其他为访问的顶点,那就就更新它的最短路径和长度dis[i].value = dis[temp].value + arc[temp][i];dis[i].path = dis[temp].path + "-->v" + to_string(i + 1);}}}}void Graph_DG::print_path(int begin) {string str;str = "v" + to_string(begin);cout << "以"<"为起点的图的最短路径为:" << endl;for (int i = 0; i != this->vexnum; i++) {if(dis[i].value!=INT_MAX)cout << dis[i].path << "=" << dis[i].value << endl;else {cout << dis[i].path << "是无最短路径的" << endl;}}}
main.cpp文件的代码 
//检验输入边数和顶点数的值是否有效,可以自己推算为啥://顶点数和边数的关系是:((Vexnum*(Vexnum - 1)) / 2) < edgebool check(int Vexnum, int edge) {if (Vexnum <= 0 || edge <= 0 || ((Vexnum*(Vexnum - 1)) / 2) < edge)return false;return true;}int main() {int vexnum; int edge;cout << "输入图的顶点个数和边的条数:" << endl;cin >> vexnum >> edge;while (!check(vexnum, edge)) {cout << "输入的数值不合法,请重新输入" << endl;cin >> vexnum >> edge;}Graph_DG graph(vexnum, edge);graph.createGraph();graph.print();graph.Dijkstra(1);graph.print_path(1);system("pause");return 0;}
6 81 3 101 5 301 6 1002 3 53 4 504 6 105 6 605 4 20
输出: 

评论
