宽搜一般要手写一个队列，深搜一般是用系统栈来实现的。
DFS之连通性模型
1112. 迷宫 - AcWing题库
import java.util.*;public class Main{static int N = 110, ha, la, hb, lb, n;static char[][] g = new char[N][N];static boolean[][] st = new boolean[N][N];static int[] dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};public static boolean dfs(int start, int end){if(g[start][end] == '#') return false;//如果一开始起点就是障碍物if(start == hb && end == lb) return true;//如果遍历到终点st[start][end] = true;//标记为已搜过for(int i = 0; i < 4; i ++){int a = start + dx[i];int b = end + dy[i];if (a < 0 || a >= n || b < 0 || b >= n) continue;//越界if(!st[a][b]){//如果这个点没走过if(dfs(a, b)) return true;}}return false;}public static void main(String[] args){Scanner sc = new Scanner(System.in);int T = sc.nextInt();while(T -- > 0){n = sc.nextInt();for(int i = 0; i < n; i ++){String s = sc.next();for(int j = 0; j < n; j ++){g[i][j] = s.charAt(j);st[i][j] = false;//由于有多组数组，所以每一次都要重置为false}}ha = sc.nextInt();la = sc.nextInt();hb = sc.nextInt();lb = sc.nextInt();if(dfs(ha, la)) System.out.println("YES");else System.out.println("NO");}}
}
1113. 红与黑 - AcWing题库
import java.util.*;public class Main{static int N = 25;static int n, m, cnt;static char[][] g = new char[N][N];static boolean[][] st = new boolean[N][N];static int[] dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};public static void dfs(int x, int y){st[x][y] = true;cnt ++;for(int i = 0; i < 4; i ++){int a = x + dx[i];int b = y + dy[i];if(a < 0 || b < 0 || a >= n || b >= m) continue;if(st[a][b]) continue;if(g[a][b] == '#') continue;dfs(a, b);}}public static void main(String[] args){Scanner sc = new Scanner(System.in);while(true){m = sc.nextInt();n = sc.nextInt();if(m == 0 && n == 0) break;int x = 0;int y = 0;cnt = 0;for(int i = 0; i < n; i ++){String s = sc.next();for(int j = 0; j < m; j ++){g[i][j] = s.charAt(j);st[i][j] = false;//由于有多组测试数组，所以每次要重置st数组if(g[i][j] == '@'){//用来确定起点的位置x = i;y = j;}}}dfs(x, y);System.out.println(cnt);}}
}
DFS之搜索顺序
1116. 马走日 - AcWing题库
import java.util.*;public class Main{static int N = 15;static int n, m, res, sx, sy;static int[] dx = {-2, -1, 1, 2, 2, 1, -1, -2};static int[] dy = {1, 2, 2, 1, -1, -2, -2, -1};static boolean[][] st = new boolean[N][N];public static void dfs(int x, int y, int cnt){if(cnt == n * m){res ++;//方案数加1return;//每次要记得返回}st[x][y] = true;for(int i = 0; i < 8; i ++){int a = x + dx[i];int b = y + dy[i];if(a < 0 || b < 0 || a >= n || b >= m) continue;if(st[a][b]) continue;dfs(a, b, cnt + 1);}st[x][y] = false;//回溯}public static void main(String[] args){Scanner sc = new Scanner(System.in);int t = sc.nextInt();while(t -- > 0){n = sc.nextInt();m = sc.nextInt();sx = sc.nextInt();sy = sc.nextInt();res = 0;//因为有多组测试数据，所以每次答案要重置dfs(sx, sy, 1);//1表示已经遍历了几个点System.out.println(res);}}
}1117. 单词接龙 - AcWing题库
import java.util.*;public class Main{static int N = 25, n, res;static int[][] g = new int[N][N];//g[n][m]表示编号为n和m的单词的最短重合的长度static int[] used = new int[N];//表示这个编号的单词用了多少次static String[] word = new String[N];//给出的单词static char lead;public static void dfs(String dragon, int last){res = Math.max(res, (int)dragon.length());//每次都比较一下used[last] ++;//每次使用都加一下for(int i = 0; i < n; i ++){if(g[last][i] != 0 && used[i] < 2){dfs(dragon + word[i].substring(g[last][i]), i);}}used[last] --;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();for(int i = 0; i < n; i ++){word[i] = sc.next();}lead = sc.next().charAt(0);//预处理出g数组for(int i = 0; i < n; i ++){for(int j = 0; j < n; j ++){String a = word[i], b = word[j];//由于要想龙的长度最长，所以要使得两个单词重叠部分最短,至少重叠一个字母for(int k = 1; k < Math.min(a.length(), b.length()); k ++){if(a.substring(a.length() - k).equals(b.substring(0, k))){g[i][j] = k;break;}}}}//遍历所有是从龙头开始的单词for(int i = 0; i < n; i ++){if(word[i].charAt(0) == lead) dfs(word[i], i);//i表示单词的编号}System.out.print(res);}
}
1118. 分成互质组 - AcWing题库
import java.util.*;public class Main{static int N = 15;static int[] q = new int[N];static int[][] groud = new int[N][N];//每一组中第几个数的下标static boolean[] st = new boolean[N];//用来判断是否用过了static int n, res = N;//求最大公因数public static int gcd(int a, int b){return (b != 0 ? gcd(b, a % b) : a);}//判断一个数与这个组内的数是否都互质public static boolean check(int[] groud, int gc, int t){//第几组，这一组有几个数，要比较的数的下标for(int i = 0; i < gc; i ++){if(gcd(q[groud[i]], q[t]) > 1) return false;//公因数大于1，不互质}return true;}public static void dfs(int g, int gc, int tc, int start){if(g >= res) return;//如果组数已经大于等于我们最小组数，就直接返回if(tc == n) res = g;//用来判断是否要开一个新的组boolean flag = true;for(int i = start; i < n; i ++){if(!st[i] && check(groud[g], gc, i)){flag = false;//有数组可以放就不用开新的数组st[i] = true;groud[g][gc] = i;dfs(g, gc + 1, tc + 1, i);st[i] = false;//回溯}}if(flag) dfs(g + 1, 0, tc, 0);//重开一个组，组内的数一开始应该为0，tc总数不变，应该从0开始搜}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();for(int i = 0; i < n; i ++){q[i] = sc.nextInt();}dfs(1, 0, 0, 0);//第一组，当前第一组中有0个数，一共已经搜索了0个数，从第0个数开始搜System.out.print(res);}
}
)
