解题思路:
注意事项:
参考代码:
package 练习; import java.util.*; public class Main { static Scanner sc = new Scanner(System.in); static int n = sc.nextInt(); static int count = 0; static ArrayList<Integer> ans = new ArrayList<>(); public static void main(String[] args) { // int[] north ={2,4,3,4}; // int[] west = {4,3,3,3}; int[] north = new int[n]; int[] west = new int[n]; for (int i = 0; i < north.length; i++) { north[i] = sc.nextInt(); } for (int i = 0; i < west.length; i++) { west[i] = sc.nextInt(); } int[][] map = new int[n][n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { map[i][j] = count++; } } dfs(map, 0, 0, north, west); } private static void dfs(int[][] map, int x, int y, int[] north, int[] west) { boolean check = true; if (x >= n || y >= n || x < 0 || y < 0) { return; } for (int i = 0; i < n - 1; i++) { if (north[i] != 0 || west[i] != 0) { check = false; } } if (north[n - 1] != 1 || west[n - 1] != 1) { check = false; } for (int i = 0; i < n; i++) { if (north[i] < 0 || west[i] < 0) { return; } } if (x == n - 1 && y == n - 1 && check == false) { return; } if (x == n - 1 && y == n - 1 && check) { ans.add(map[x][y]); for (int i = 0; i < ans.size(); i++) { System.out.print(ans.get(i) + " "); } System.exit(0); } west[x]--; north[y]--; ans.add(map[x][y]); dfs(map, x, y + 1, north, west); dfs(map, x + 1, y, north, west); dfs(map, x - 1, y, north, west); dfs(map, x, y - 1, north, west); ans.remove(ans.size()-1); west[x]++; north[y]++; } }
0.0分
0 人评分
C语言程序设计教程(第三版)课后习题11.5 (C语言代码)浏览:932 |
C语言程序设计教程(第三版)课后习题5.4 (C语言代码)浏览:1327 |
【偶数求和】 (C语言代码)浏览:588 |
【蟠桃记】 (C语言代码)浏览:1084 |
C语言程序设计教程(第三版)课后习题5.6 (C语言代码)浏览:580 |
C语言程序设计教程(第三版)课后习题9.4 (C语言代码)浏览:699 |
勾股数 (C语言代码)浏览:830 |
字符串比较 (C语言代码)浏览:770 |
C语言程序设计教程(第三版)课后习题7.5 (C语言代码)浏览:592 |
C语言程序设计教程(第三版)课后习题10.1 (C语言代码)浏览:826 |