DFS、BFS的定義及C語言算法實(shí)現(xiàn)請(qǐng)參照
連通圖遍歷策略之深度優(yōu)先搜索（C語言）
連通圖遍歷策略之廣度優(yōu)先搜索（C語言）

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為了便于理解這里的數(shù)據(jù)是一個(gè)無向圖，要求輸出遍歷順序

下面只給出用例和算法，之后可以根據(jù)后面的三個(gè)題目進(jìn)行深入學(xué)習(xí)
Input:
5 5
1 2
1 3
1 5
2 4
3 5
Output:
1 2 4 3 5

DFS

import java.util.Scanner;public class DFS {static int[][] e = new int[100][100];static int[] book = new int[100];static int n, m;static int sum = 0;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();e[a][b] = 1;e[b][a] = 1;}book[1] = 1;dfs(1);}public static void dfs(int cur) {System.out.print(cur + " ");sum++;if (sum == n) {return;}for (int i = 1; i <= n; i++) {if (e[cur][i] == 1 && book[i] == 0) {book[i] = 1;dfs(i);}}return;} }

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BFS

import java.util.LinkedList; import java.util.Queue; import java.util.Scanner;public class BFS {static int[][] e = new int[100][100];static int[] book = new int[100];static int n, m;static Queue<Integer> queue = new LinkedList<>();static Scanner input = new Scanner(System.in);public static void main(String[] args) {int a, b;n = input.nextInt();m = input.nextInt();for (int i = 0; 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++) {a = input.nextInt();b = input.nextInt();e[a][b] = 1;e[b][a] = 1;}queue.offer(1);book[1] = 1;bfs();}public static void bfs() {while (!queue.isEmpty()) {int cur = queue.peek();for (int i = 1; i <= n; i++) {if (e[cur][i] == 1 && book[i] == 0) {book[i] = 1;queue.offer(i);}}System.out.print(queue.remove() + " ");}return;} }

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下面是一個(gè)全排列的簡(jiǎn)單題目，也用到了dfs，屬于偏簡(jiǎn)單的一道題目,用來給后面的題目做鋪墊

問題描述：假設(shè)有編號(hào)為1、2、3的三張卡片和編號(hào)1、2、3的三個(gè)盒子，現(xiàn)需將這三張撲克牌分別放到三個(gè)盒子里面，并且每個(gè)盒子只能放一張撲克牌，請(qǐng)問一共有幾種不同的方法？
Output：
123 132 213 231 312 321

import java.util.Scanner; /*** DFS算法基本模型：* void dfs(int step) {* 判斷邊界* 嘗試每一種可能 for {* 繼續(xù)下一步def(step + 1)* }* 返回* }* */public class DFS {static int n;static int[] book = new int[10];static int[] a = new int[10];public static void main(String[] args) {Scanner input = new Scanner(System.in);n = input.nextInt();dfs(1);}public static void dfs(int step) {if (step == n + 1) {/*** 打印，并返回* */for (int i = 1; i <= n; i++) {System.out.print(a[i] + " ");}System.out.println();return;}for (int i = 0; i < n; i++) {/*** 標(biāo)記數(shù)組book，用來檢測(cè)盒子中是否已經(jīng)放入* */if (book[i] == 0) {/*** 將i號(hào)撲克牌放入到第step個(gè)盒子中* */a[step] = i;book[i] = 1;/*** 處理step+1個(gè)小盒子，注意要記得將剛才嘗試的撲克牌收回* */dfs(step + 1);book[i] = 0;}}return;}}

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