迷宫般的回溯

Posted 2023-02-22

【中文标题】迷宫般的回溯

【英文标题】：Maze-like backtracking

【发布时间】：2017-12-21 09:38:12

【问题描述】：

所以，我正在尝试使用给定的开始和结束位置 (X, Y) 坐标制作一个迷宫求解器，但我有一个条件：每次从一个点到另一个点时，它都应该检查新位置是否为低于前一个 (a[x][y]

#include <stdio.h>

#define N 4

int a[N][N] = 

    35, 75, 80, 12, 
    13, 12, 11, 3, 
    32, 9, 10, 8, 
    12, 2, 85, 1
;

int sol[N][N];

int h, k, count;
int end_x = -1, end_y = -1;
int start_x = -1, start_y = -1;

void print()

    int i, j;

    for (i = 0; i < N; i++)
    
        for (j = 0; j < N; j++)
        
            if (sol[i][j] > 0)
                printf("%d ", sol[i][j]);

            else
                printf("_ ");
        
        printf("\n");
    

    printf("\n");


int solution(int x, int y)

    return (x == end_x && y == end_y);


int valid(int x, int y)

    return (x >= 0 && x < N && y >= 0 && y < N && a[x][y] <= h && !sol[x][y]);


void back(int x, int y)

    if (valid(x, y))
    
        k ++;
        sol[x][y] = k;
        h = a[x][y]; // right here I'm updating the variabile

        if (solution(x, y))
        
            count ++;
            print();
        
        else
        
            back(x + 1, y);
            back(x, y + 1);
            back(x - 1, y);
            back(x, y - 1);
        

        sol[x][y] = 0;
        h = a[x][y]; // I actually don't know where to put this
        k --;
    


int main(void)

    int i, j;

    for (i = 0; i < N; i++)
    
        for (j = 0; j < N; j++)
            printf("%d ", a[i][j]);

        printf("\n");
    

    while (start_x >= N || start_x < 0 || start_y >= N || start_y < 0)
    
        printf(">> Start X: "); scanf("%d", &start_x);
        printf(">> Start Y: "); scanf("%d", &start_y);
    

    h = a[start_x][start_y];

    while (end_x >= N || end_x < 0 || end_y >= N || end_y < 0)
    
        printf(">> End X: "); scanf("%d", &end_x);
        printf(">> End Y: "); scanf("%d", &end_y);
    

    printf("Generated solutions:\n");
    back(start_x, start_y);

    if (!count)
        printf("No path was found!\n");

    return 0;

因此，对于start_x = 0、start_y = 0、end_x = 1、end_y = 3，它应该带来 35 -> 13 -> 12 -> 11 -> 3 和 35 - > 13 -> 12-> 11 -> 10 -> 8 -> 3 个解决方案。

如果没有这个条件，算法就可以正常工作，只是我不知道在哪里更新 h 变量。

【参考方案1】：

我认为你是在正确的轨道上。但是变量h 似乎是多余的。相反，您还需要两件事：

1) 您之前是否访问过特定的单元格。

2) 您之前移动的值，因此当您移动到下一个单元格或考虑移动到下一个单元格时，您要验证它是否有效。

你修改后的代码如下：

#include <stdio.h>

#define N 4

int a[N][N] =

    35, 75, 80, 12,
    13, 12, 11, 3,
    32, 9, 10, 8,
    12, 2, 85, 1
;

int sol[N][N];
int visited[N][N];

int h, k, count;
int end_x = -1, end_y = -1;
int start_x = -1, start_y = -1;

void print()

    int i, j;

    for (i = 0; i < N; i++)
    
        for (j = 0; j < N; j++)
        
            if (sol[i][j] > 0)
                printf("%d ", sol[i][j]);

            else
                printf("_ ");
        
        printf("\n");
    

    printf("\n");


int solution(int x, int y)

    return (x == end_x && y == end_y);


int valid(int x, int y, int currentCellValue)

    return (x >= 0 && x < N && y >= 0 && y < N && a[x][y] <= currentCellValue && !sol[x][y] && !visited[x][y]);


void back(int x, int y, int curr)

    if (valid(x, y, curr))
    
        k ++;
        sol[x][y] = k;
        visited[x][y] = 1;

        if (solution(x, y))
        
            count ++;
            print();
        
        else
        
            back(x + 1, y, a[x][y]);
            back(x, y + 1, a[x][y]);
            back(x - 1, y, a[x][y]);
            back(x, y - 1, a[x][y]);
        

        sol[x][y] = 0;
        visited[x][y] = 0;
        k --;
    


int main(void)

    int i, j;

    for (i = 0; i < N; i++)
    
        for (j = 0; j < N; j++)
            printf("%d ", a[i][j]);

        printf("\n");
    

    while (start_x >= N || start_x < 0 || start_y >= N || start_y < 0)
    
        printf(">> Start X: "); scanf("%d", &start_x);
        printf(">> Start Y: "); scanf("%d", &start_y);
    

    h = a[start_x][start_y];

    while (end_x >= N || end_x < 0 || end_y >= N || end_y < 0)
    
        printf(">> End X: "); scanf("%d", &end_x);
        printf(">> End Y: "); scanf("%d", &end_y);
    

    printf("Generated solutions:\n");
    back(start_x, start_y, a[start_x][start_y]);

    if (!count)
        printf("No path was found!\n");

    return 0;

在调用back 期间，请注意我如何传递当前单元格的值，该值被传递给isValid 函数以确保移动是否有效。

int visited[N][N]; 确保您不会一遍又一遍地访问同一个单元格。

【讨论】：

是的，我正在考虑提出一个新的论点，但我并不确定。谢谢！