Blue Bridge Cup DFS find the shortest path (simple)

Problem Description

As shown in the figure, there are 5 locations on the Baidu map, and each location is a one-way path. Try to find the shortest path from 1 to 5.

A 5*5 two-dimensional matrix can be obtained from the figure, and the shortest path can be found by using the depth search algorithm.

Reference Code

import java.util.ArrayList;
import java.util.Arrays;

public class 迪杰斯特拉 {
    //根据路径图写出连通图.0为不连通,若联通记录他的权值
	static int[][] map = {
		{0,2,0,0,10},
		{0,0,3,0,7},
		{4,0,0,4,0},
		{0,0,0,0,5},
		{0,0,3,0,0}
	};
    //顶点
	static int[] nodes = {1,2,3,4,5};

	
	public static void main(String[] args) {
        //用来记录此顶点是否标记为确定最短路径的顶点
		boolean check[] = new boolean[nodes.length];
        //用来记录从0到此顶点所用的最短路径
		int[] dist = new int[nodes.length];
        //用来记录到达他的最近顶点下标
		int[] path = new int[nodes.length];
		//目标定点下表,(5)下表为4
		int end = 4;
		//首先遍历0到所有顶点是否联通
		for (int i = 0; i < map.length; i++){
			//如果联通,dist记录他的权值,path记录到达他的最近顶点为0
			if (map[0][i] != 0) {
				dist[i] = map[0][i];
				path[i] = 0;
			}else {//不连通,dist为最大值(为了方便后面更新他的权值),path为-1(为了后面更新录到达他的最近的顶点)
				dist[i] = Integer.MAX_VALUE;
				path[i] = -1;
			}
		}
		//循环条件:end定点为被划分为确定最短路径的顶点
		while (!check[end]) {
			int min = Integer.MAX_VALUE;
			int nodes = 0;
			//从没有确定最小值的顶点选择一个最小距离
			for (int i = 0; i < check.length; i++) {
				if (!check[i] && dist[i] < min) {
					min = dist[i];
					nodes = i;
				}
			}
			//讲该顶点标记为已找到到达他最近的顶点
			check[nodes] = true;
			//查询最小顶点到附近顶点最短路径
			for (int i = 0; i < map.length; i++) {
				if (map[nodes][i]!=0) {//确保是通路状态。
						//0-当前最短距离+当前顶点到附近顶点距离<0-附近顶点距离
					if (dist[nodes] + map[nodes][i] < dist[i]) {
						dist[i] = dist[nodes] + map[nodes][i];
						path[i] = nodes;//更新到这个节点的最短顶点Path
					}
				}
			}
		}
		//输出从0到目标定点所走的最短路径
		//根据path倒序输出
		int index = end;
		System.out.print(index+" ");
		while (index != 0) {
			System.out.print(path[index]+" ");
			index = path[index];
		}
	}
}