最短路径–城市路径问题：

问题描述：求从1号城市到5号城市的最短路径长度

Input：
5 8
1 2 2
1 5 10
2 3 3
2 5 7
3 1 4
3 4 4
4 5 5
5 3 3
Output：
9

DFS

import java.util.Scanner;
public class minPath {static int min = 99999999;static int[][] e = new int[100][100];static int[] book = new int[100];static int n, m;static Scanner input = new Scanner(System.in);public static void main(String[] args) {n = input.nextInt();m = input.nextInt();for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {if (i == j) {e[i][j] = 0;} else {e[i][j] = 99999999;}}}for (int i = 1; i <= m; i++) {int a = input.nextInt();int b = input.nextInt();int c = input.nextInt();e[a][b] = c;}book[1] = 1;dfs(1, 0);System.out.println(min);}public static void dfs(int cur, int dis) {/*** 如果当前路径大于之前找到的最小值，可直接返回* */if (dis > min) {return;}/*** 判断是否达到最后一个结点，更新最小值，返回* */if(cur == n) {if (dis < min) {min = dis;return;}}/*** 当前点到其他各点之间可连通但是还未添加进来时，遍历执行* */for (int i = 1; i <= n; i++) {if (e[cur][i] != 99999999 && book[i] == 0) {book[i] = 1;dfs(i, dis+e[cur][i]);/*** 回溯**/book[i] = 0;}}return;}
}

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最短路径–转乘问题：

问题描述：求从1号城市到5号城市的最短路径长度
Input：
5 7 1 5
1 2
1 3
2 3
2 4
3 4
3 5
4 5
Output：
2

这个题目与上一个的区别就在于，这些路径是没有权值的，也就是说只需要找出到达一个点的最短距离就可以了，记录次数即可。

BFS

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;class node {int x;int s;node(int x, int s) {this.x = x;this.s = s;}
}public class minPath {static int[][] e = new int[51][51];static int[] book = new int[51];static int n, m;static int start, end;static int mark, sum;static Queue<node> queue = new LinkedList<>();static Scanner input = new Scanner(System.in);public static void main(String[] args) {n = input.nextInt();m = input.nextInt();start = input.nextInt();end = input.nextInt();for (int i = 1; i <= n; i++) {for (int j = 0; j <= m; j++) {if (i == j) {e[i][j] = 0;} else {e[i][j] = 99999999;}}}for (int i = 1; i <= m; i++) {int a = input.nextInt();int b = input.nextInt();e[a][b] = 1;e[b][a] = 1;}queue.offer(new node(start, 0));book[1] = start;bfs();System.out.println(sum);}public static void bfs() {int flag = 0;while (!queue.isEmpty()) {int cur = queue.peek().x;for (int i = 1; i <= n; i++) {if(e[cur][i] != 99999999 && book[i] == 0) {mark = i;sum = queue.peek().s + 1;queue.offer(new node(i, sum));book[i] = 1;}if(mark == end) {flag = 1;break;}}if(flag == 1) {break;}queue.remove();}return;}
}

基本上能用深度优先的问题都可以用广度优先，但是广度优先更适合无向图，或者说所有边的权值一样的情况，大家可以通过多做题灵活使用这两种方法