南极人探险 2022-01-24 22:40 采纳率: 87.5%
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已结题

尝试用java解决迷宫问题。但是失败了,希望能找出原因

下图给出了一个迷宫的平面图,其中标记为 1的为障碍,标记为 0的为可以通行的地方。

010000
000100
001001
110000

其中 D、U、L、R 分别表示向下、向上、向左、向右走。 对于下面这个更复杂的迷宫(30行 50列),请找出一种通过迷宫的方式,其使用的步数最少,在步数最少的前提下,请找出字典序最小的一个作为答案。

01010101001011001001010110010110100100001000101010
00001000100000101010010000100000001001100110100101
01111011010010001000001101001011100011000000010000
01000000001010100011010000101000001010101011001011
00011111000000101000010010100010100000101100000000
11001000110101000010101100011010011010101011110111
00011011010101001001001010000001000101001110000000
10100000101000100110101010111110011000010000111010
00111000001010100001100010000001000101001100001001
11000110100001110010001001010101010101010001101000
00010000100100000101001010101110100010101010000101
11100100101001001000010000010101010100100100010100
00000010000000101011001111010001100000101010100011
10101010011100001000011000010110011110110100001000
10101010100001101010100101000010100000111011101001
10000000101100010000101100101101001011100000000100
10101001000000010100100001000100000100011110101001
00101001010101101001010100011010101101110000110101
11001010000100001100000010100101000001000111000010
00001000110000110101101000000100101001001000011101
10100101000101000000001110110010110101101010100001
00101000010000110101010000100010001001000100010101
10100001000110010001000010101001010101011111010010
00000100101000000110010100101001000001000000000010
11010000001001110111001001000011101001011011101000
00000110100010001000100000001000011101000000110011
10101000101000100010001111100010101001010000001000
10000010100101001010110000000100101010001011101000
00111100001000010000000110111000000001000000001011
10000001100111010111010001000110111010101101111000

请注意在字典序中 D<L<R<U。

我的题解:

package 蓝桥;
import java.util.*;
public class exercise6 {
    static void bfs(char[][] map, Stack set, String goal,Stack step) {
        if(set.contains(goal)) {
            int len=step.size();
            while(len>0) {
                System.out.print(step.pop());
                len--;
            }
        }
        //3、循环区
        String[] temp = ((String)set.peek()).split(",");
        int y = Integer.parseInt(temp[0]);
        int x = Integer.parseInt(temp[1]);  //将Set from中的一组xy取出来
            
        if(y > 0 && map[y-1][x] == '0') {   //上方
             map[y-1][x] = '1';  //做好标记,说明已经搜索过
             set.push(y-1+","+x); //将其加入下一轮迭代的Set from中
             step.push("U");
           }
            
        else if(y < map.length-2 && map[y+1][x] == '0') {    //下方
             map[y+1][x] = '1';
             set.add(y+1+","+x);
             step.add("D");
          }
            
        else if(x > 0 && map[x-1][y] == '0') {   //左方
             map[y][x-1] = '1';
             set.push(y+","+(x-1));
             step.push("L");
         }
            
        else if(x < map[0].length-2 && map[x+1][y] == '0') { //右方
            map[y][x+1] = '1';
            set.push(y+","+x+1);
            step.push("R");
         }else {
             map[y][x]='1';
             if(set.isEmpty())
                 return;
             set.pop();
             step.pop();
         }
        //4、处理区
        bfs(map, set, goal,step);
        
       
    }
    
    public static void main(String[]args)throws Exception {
        Scanner in=new Scanner(System.in);
        char [][]map=new char[30][50];
        for(int i=0;i<30;i++) {
            map[i]=in.nextLine().trim().toCharArray();
        }
        //迷宫地图已经布置完成
        Stack s=new Stack();
        Stack step=new Stack();
        s.push(0+","+0);
        bfs(map, s, 29+","+49,step);
        
    }

}


img

  • 写回答

1条回答 默认 最新

  • zlebhs 2022-01-25 13:57
    关注

    取了个bfs的名字,内容却是dfs。迷宫出入口是哪里?左上和右下么?
    先来简单的分析一下题目,做迷宫题无非就是dfs或者bfs,看题目一共有2个要求:

    1. 步数要最小:如何保证步数最小?
      dfs:可以把所有路径都求出来,然后找到最小的步数。
      bfs:本身每次访问的下一层,都是最小路径,不可能会绕路,同求树高之类的方法
    2. 字典序要最小:这个比较容易,无论dfs、bfs只要按照DLRU这个顺序进行访问即可。

    综上来看,采用bfs的方式,更适合作为本题的解。更何况数据量比较大,如果不做剪枝、记忆之类的方法,dfs肯定有溢出风险,就像你的错误那样。
    bfs:

    public class Demo {
        // 4个方向,按照DLRU顺序
        private int[][] dires = {{1, 0}, {0, -1}, {0, 1}, {-1, 0}};
        // 方向对应的英文
        private String[] direStr = {"D", "L", "R", "U"};
    
        public String bfs(int[][] map) {
            int n = map.length, m = map[0].length;
            // 记录每个将要访问的点
            Queue<int[]> points = new LinkedList<>();
            points.add(new int[]{0, 0});
            // 记录每个将要访问的点的路径
            Queue<String> paths = new LinkedList<>();
            paths.add("");
            // 常规bfs模板
            while (!points.isEmpty()) {
                // 两个队列同进同出
                int[] point = points.poll();
                String path = paths.poll();
                // 达到终点
                if (point[0] == n - 1 && point[1] == m - 1) return path;
                // 将访问过的点标记为2
                map[point[0]][point[1]] = 2;
                // 遍历4个方向
                for (int i = 0; i < dires.length; i++) {
                    int r = point[0] + dires[i][0];
                    int c = point[1] + dires[i][1];
                    // 判断能否访问
                    if (r >= 0 && c >= 0 && r < n && c < m && map[r][c] == 0) {
                        points.add(new int[]{r, c});
                        paths.add(path + direStr[i]);
                    }
                }
            }
            return "";
        }
    
        public static void main(String[] args) {
            String str = "01010101001011001001010110010110100100001000101010"
                    + "00001000100000101010010000100000001001100110100101"
                    + "01111011010010001000001101001011100011000000010000"
                    + "01000000001010100011010000101000001010101011001011"
                    + "00011111000000101000010010100010100000101100000000"
                    + "11001000110101000010101100011010011010101011110111"
                    + "00011011010101001001001010000001000101001110000000"
                    + "10100000101000100110101010111110011000010000111010"
                    + "00111000001010100001100010000001000101001100001001"
                    + "11000110100001110010001001010101010101010001101000"
                    + "00010000100100000101001010101110100010101010000101"
                    + "11100100101001001000010000010101010100100100010100"
                    + "00000010000000101011001111010001100000101010100011"
                    + "10101010011100001000011000010110011110110100001000"
                    + "10101010100001101010100101000010100000111011101001"
                    + "10000000101100010000101100101101001011100000000100"
                    + "10101001000000010100100001000100000100011110101001"
                    + "00101001010101101001010100011010101101110000110101"
                    + "11001010000100001100000010100101000001000111000010"
                    + "00001000110000110101101000000100101001001000011101"
                    + "10100101000101000000001110110010110101101010100001"
                    + "00101000010000110101010000100010001001000100010101"
                    + "10100001000110010001000010101001010101011111010010"
                    + "00000100101000000110010100101001000001000000000010"
                    + "11010000001001110111001001000011101001011011101000"
                    + "00000110100010001000100000001000011101000000110011"
                    + "10101000101000100010001111100010101001010000001000"
                    + "10000010100101001010110000000100101010001011101000"
                    + "00111100001000010000000110111000000001000000001011"
                    + "10000001100111010111010001000110111010101101111000";
            
            int[][] map = new int[30][50];
            for (int i = 0; i < 30; i++) {
                for (int j = 0; j < 50; j++) {
                    map[i][j] = str.charAt(i * 50 + j) - '0';
                }
            }
            System.out.println(new Demo().bfs(map));
    
        }
    
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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