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算法模板（Java版）_DFS与BFS

news 2025/9/6 8:17:58

@ZZHow(ZZHow1024)

搜索方式	数据结构	空间	特点
DFS	Steak	$O (h)$	不具有最短路
BFS	Queue	$O(2 ^ h)$	最短路

DFS

DFS（深度优先搜索），回溯时记得回复现场。

必要时进行剪枝操作。

例题：AcWing 842. 排列数字

import java.util.Scanner;public class Main {static int n;static int[] path;static boolean[] st;public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();path = new int[n + 1];st = new boolean[n + 1];dfs(0);}public static void dfs(int u) {if (u == n) {for (int i = 0;i < n; i++)System.out.print(path[i] + " ");System.out.println();return;}for (int i = 1; i <= n; i++) {if (!st[i]) {st[i] = true;path[u] = i;dfs(u + 1);st[i] = false;}}}
}

例题：AcWing 843. n-皇后问题

按行搜索

import java.util.Scanner;public class Main {static final int N = 20;static int n;static char[][] g = new char[N][N];static boolean[] col = new boolean[N];static boolean[] dg = new boolean[N];static boolean[] udg = new boolean[N];public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)g[i][j] = '.';dfs(0);}public static void dfs(int u) {// 搜索出一组答案if (u == n) {for (int i = 0; i < n; i++){for (int j = 0; j < n; j++)System.out.print(g[i][j]);System.out.println();}System.out.println();}// 枚举一行可以放棋子的位置for (int i = 0; i < n; i++) {if (!col[i] && !dg[u + i] && !udg[n - u + i]) {g[u][i] = 'Q';col[i] = dg[u + i] = udg[n - u + i] = true;dfs(u + 1);col[i] = dg[u + i] = udg[n - u + i] = false;g[u][i] = '.';}}}
}

按格子依次搜索

import java.util.Scanner;public class Main {static final int N = 20;static int n;static char[][] g = new char[N][N];static boolean[] row = new boolean[N];static boolean[] col = new boolean[N];static boolean[] dg = new boolean[N];static boolean[] udg = new boolean[N];public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)g[i][j] = '.';dfs(0, 0, 0);}public static void dfs(int x, int y, int s) {// 行出界if (y == n) {y = 0;x++;}// 搜索出一组答案if (x == n) {if (s == n) {for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++)System.out.print(g[i][j]);System.out.println();}System.out.println();}return;}// 不放皇后dfs(x, y + 1, s);// 放皇后if (!row[x] && !col[y] && !dg[x + y] && !udg[x - y + n]) {g[x][y] = 'Q';row[x] = col[y] = dg[x + y] = udg[x - y + n] = true;dfs(x, y + 1, s + 1);row[x] = col[y] = dg[x + y] = udg[x - y + n] = false;g[x][y] = '.';}}
}

BFS

BFS（宽度优先搜索），使用队列。

例题：AcWing 844. 走迷宫

import java.util.Scanner;public class Main {static final int N = 110;static int n;static int m;static int[][] g = new int[N][N];static int[][] d = new int[N][N];static Pair[] q = new Pair[N * N];public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();m = scanner.nextInt();for (int i = 0; i < n; i++)for (int j = 0; j < m; j++)g[i][j] = scanner.nextInt();System.out.println(bfs());}public static int bfs() {// 队头int hh = 0;// 队尾int tt = 0;q[0] = new Pair(0, 0);for (int i = 0; i < d.length; i++)for (int j = 0; j < d[i].length; j++)d[i][j] = -1;d[0][0] = 0;int[] dx = {-1, 0, 1, 0};int[] dy = {0, 1, 0, -1};while (hh <= tt) {Pair t = q[hh++]; // 取出队头for (int i = 0; i < 4; i++) {int x = t.first + dx[i];int y = t.second + dy[i];if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 && d[x][y] == -1) {d[x][y] = d[t.first][t.second] + 1;q[++tt] = new Pair(x, y);}}}return d[n - 1][m - 1];}
}class Pair {public int first;public int second;public Pair(int first, int second) {this.first = first;this.second = second;}
}

树与图的深度优先遍历

树是一种特殊的图（无环连通图）

有向图的存储方法

邻接矩阵
邻接表

例题：AcWing 846. 树的重心

import java.util.Scanner;public class Main {static final int N = 100010;static final int M = N * 2;static int n;static int[] h = new int[N];static int[] e = new int[M];static int[] ne = new int[M];static boolean[] st = new boolean[N];static int idx = 0;static int ans = N;public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();for (int i = 0; i < h.length; i++)h[i] = -1;for (int i = 0; i < n - 1; i++) {int a = scanner.nextInt();int b = scanner.nextInt();add(a, b);add(b, a);}bfs(1);System.out.println(ans);}public static void add(int a, int b) {e[idx] = b;ne[idx] = h[a];h[a] = idx++;}public static int bfs(int u) {st[u] = true;int sum = 1; // 当前子树的大小int res = 0; // 删除当前点后连通块的最大值for (int i = h[u]; i != -1; i = ne[i]) {int j = e[i];if (!st[j]) {int s = bfs(j);res = Math.max(res, s);sum += s;}}res = Math.max(res, n - sum);ans = Math.min(ans, res);// 返回当前子树的大小return sum;}
}

树与图的广度优先遍历

重边：两个点之间有两条边

自环：一条边指向自己

例题：AcWing 847. 图中点的层次

import java.util.Scanner;public class Main {static final int N = 100010;static int n;static int m;static int[] h = new int[N];static int[] e = new int[N];static int[] ne = new int[N];static int idx = 0;static int[] q = new int[N];static int[] d = new int[N];public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();m = scanner.nextInt();for (int i = 0; i <= n; i++)h[i] = -1;while (m-- != 0) {int a = scanner.nextInt();int b = scanner.nextInt();add(a, b);}System.out.println(bfs());}public static void add(int a, int b) {e[idx] = b;ne[idx] = h[a];h[a] = idx++;}public static int bfs() {int hh = 0;int tt = 0;for (int i = 0; i <= n; i++)d[i] = -1;q[0] = 1;d[1] = 0;while (hh <= tt) {int t = q[hh++];for (int i = h[t]; i != -1; i = ne[i]) {int j = e[i];if (d[j] == -1) {d[j] = d[t] + 1;q[++tt] = j;}}}return d[n];}
}

有向图的拓扑序列

例题：AcWing 848. 有向图的拓扑序列

import java.util.Scanner;public class Main {static final int N = 100010;static int n;static int m;static int[] h = new int[N];static int[] e = new int[N];static int[] ne = new int[N];static int idx = 0;static int[] q = new int[N];static int[] d = new int [N];public static void main(String[] args) {Scanner scanner = new Scanner(System.in);n = scanner.nextInt();m = scanner.nextInt();for (int i = 0; i <= n; i++)h[i] = -1;while (m-- != 0) {int a = scanner.nextInt();int b = scanner.nextInt();add(a, b);d[b]++;}if (topSort()) {for (int i = 0; i < n; i++)System.out.print(q[i] + " ");} else {System.out.println("-1");}}public static void add(int a, int b) {e[idx] = b;ne[idx] = h[a];h[a] = idx++;}public static boolean topSort() {int hh = 0;int tt = -1;for (int i = 1; i <= n; i++)if (d[i] == 0)q[++tt] = i;while (hh <= tt) {int t = q[hh++];for (int i = h[t]; i != -1; i = ne[i]) {int j = e[i];d[j]--;if (d[j] == 0)q[++tt] = j;}}return tt == n - 1;}
}