Question

我只是通过递归对连接的组件标签进行编码但是我想要使用堆栈和结构的非递归版本。

typedef struct _node{
int Xpos;
int Ypos;
int dir;    
}node;

node *stack[MAX];
node *push(node *t) {
stack[++top] = t;
return t;
}

node *pop() {
return stack[top--];
    }

结构组件确实具有Xposition和Yposition和Direction。例如） if（t-> dir == 1）{t-> Xpos = ++ t-> Xpos，t-> Ypos = --t-> Ypos}; 8邻居标签我真的需要帮助！ //编辑

//编辑这是我通过递归所做的

#define MAP_SIZE 15
#define INILBCNT    10
int cnt = 1;
int input_map[MAP_SIZE][MAP_SIZE] =
{ {0,0,0,0,0,0,0,0,0,0,0,0,1,1,0},
{0,1,1,1,1,1,1,0,0,0,0,0,1,1,0},
{0,1,0,0,0,0,1,0,0,0,1,1,0,1,0},
{0,1,0,1,1,0,1,0,1,1,1,0,0,1,0},
{0,1,0,0,1,0,1,0,1,0,0,0,0,1,0},
{0,1,1,0,1,0,1,0,1,0,0,1,0,1,0},
{0,0,0,1,1,0,1,0,1,0,1,1,0,1,0},
{0,1,0,1,1,0,1,0,1,0,1,0,0,1,0},
{0,1,0,0,0,0,0,0,1,0,1,0,0,1,0},
{0,1,0,0,0,0,0,1,1,0,1,0,0,1,0},
{0,1,1,1,1,1,0,1,0,0,1,1,0,1,0},
{0,0,0,0,0,1,0,1,0,1,0,0,0,1,0},
{0,1,1,1,0,1,0,1,0,0,0,0,0,1,0},
{0,1,0,0,0,1,0,1,1,1,1,1,1,1,0},
{0,1,1,1,1,1,0,0,0,0,0,0,0,1,0}
};



void PrintMap()
{
system("cls");

for (int j = 0; j < MAP_SIZE; j++)
{
    for (int i = 0; i < MAP_SIZE; i++)
    {
        printf("%d ", input_map[j][i]);
    }
    printf("\n");   }  }

void glassFire(int mask_Y, int mask_X)
{

if (input_map[mask_Y][mask_X] != 1)
    return;
else
    if (input_map[mask_Y][mask_X] == 1)
    {
        input_map[mask_Y][mask_X] = cnt;            
        glassFire(mask_Y, mask_X + 1);
        glassFire(mask_Y + 1, mask_X);
        glassFire(mask_Y + 1, mask_X + 1);
        glassFire(mask_Y + 1, mask_X - 1);
        glassFire(mask_Y - 1, mask_X + 1);
        glassFire(mask_Y - 1, mask_X);
        glassFire(mask_Y - 1, mask_X + 1);
        glassFire(mask_Y, mask_X - 1); }}

void Serching()
{
PrintMap();
for (int i = 0; i < MAP_SIZE; i++)
{
    for (int j = 0; j < MAP_SIZE; j++)
    {

        if (input_map[j][i] == 1)
        {
            cnt++;
            glassFire(j, i);}
      }
   }
}

这段代码很好用。

Answer 1

这段代码很好用。

否，此代码具有未定义的行为，因为由于递归调用的添加/删除，您在input_map和/或mask_Y中读/写时mask_X和/或mask_Y的值为-1和/或15 1至mask_X / void glassFire(int mask_Y, int mask_X) { if (input_map[mask_Y][mask_X] != 1) return; else if (input_map[mask_Y][mask_X] == 1) { ... } }

在

void glassFire(int mask_Y, int mask_X)
{
   if ((mask_Y >= 0) && (mask_X >= 0) &&
       (mask_Y < MAP_SIZE) && (mask_X < MAP_SIZE) &&
       (input_map[mask_Y][mask_X] == 1))
   {
        ...
   }
}

第一个 if 没有用，但是您必须检查索引，可以是：

glassFire(mask_Y - 1, mask_X + 1);

您拨打两次glassFire(mask_Y - 1, mask_X - 1);，一次必须用丢失的glassFire(mask_Y, mask_X + 1); glassFire(mask_Y + 1, mask_X); glassFire(mask_Y + 1, mask_X + 1); glassFire(mask_Y + 1, mask_X - 1); glassFire(mask_Y - 1, mask_X + 1); glassFire(mask_Y - 1, mask_X); glassFire(mask_Y - 1, mask_X - 1); /* modified */ glassFire(mask_Y, mask_X - 1); }}代替。

而不是做

    static const YX offset[] = {
      {0,1}, {1,0}, {1,1}, {1,-1}, {-1,1}, {-1,0}, {-1,-1}, {0,-1}
    };

    for (int dir = 0; dir != 8; ++dir) 
      glassFire(mask_Y + offset[dir].y, mask_X + offset[dir].x);

您可以使用循环（这将有助于手动管理递归，减少代码量）：

typedef struct YX { int y; int x; } YX;

与void glassFire(int mask_Y, int mask_X) { YX * stack = NULL; size_t sz = 0; size_t deep = 0; for (;;) { if (input_map[mask_Y][mask_X] == 1) /* need to check, can be false now */ { input_map[mask_Y][mask_X] = cnt; static const YX offset[] = { {0,1}, {1,0}, {1,1}, {1,-1}, {-1,1}, {-1,0}, {-1,-1}, {0,-1} }; /* up to 8 positions to save */ if ((deep + 8) > sz) { sz += 8; stack = realloc(stack, sz * sizeof(YX)); } for (int dir = 0; dir != 8; ++dir) { int ny = mask_Y + offset[dir].y; int nx = mask_X + offset[dir].x; if ((ny >= 0) && (nx >= 0) && (ny < MAP_SIZE) && (nx < MAP_SIZE) && (input_map[ny][nx] == 1)) { stack[deep].y = ny; stack[deep].x = nx; deep += 1; } } } if (deep == 0) /* empty */ break; deep -= 1; mask_Y = stack[deep].y; mask_X = stack[deep].x; } free(stack); }

要删除递归调用，您的包含方向的结构是不合适的，因为递归调用可以执行其他递归调用等，您只需要保存以后要管理的索引即可。

幸运的是，结果不取决于递归调用的顺序。

解决方案可以是：

#include <stdio.h>
#include <stdlib.h>

#define MAP_SIZE 15
int cnt = 1;
int input_map[MAP_SIZE][MAP_SIZE] =
{ {0,0,0,0,0,0,0,0,0,0,0,0,1,1,0},
{0,1,1,1,1,1,1,0,0,0,0,0,1,1,0},
{0,1,0,0,0,0,1,0,0,0,1,1,0,1,0},
{0,1,0,1,1,0,1,0,1,1,1,0,0,1,0},
{0,1,0,0,1,0,1,0,1,0,0,0,0,1,0},
{0,1,1,0,1,0,1,0,1,0,0,1,0,1,0},
{0,0,0,1,1,0,1,0,1,0,1,1,0,1,0},
{0,1,0,1,1,0,1,0,1,0,1,0,0,1,0},
{0,1,0,0,0,0,0,0,1,0,1,0,0,1,0},
{0,1,0,0,0,0,0,1,1,0,1,0,0,1,0},
{0,1,1,1,1,1,0,1,0,0,1,1,0,1,0},
{0,0,0,0,0,1,0,1,0,1,0,0,0,1,0},
{0,1,1,1,0,1,0,1,0,0,0,0,0,1,0},
{0,1,0,0,0,1,0,1,1,1,1,1,1,1,0},
{0,1,1,1,1,1,0,0,0,0,0,0,0,1,0}
};

void PrintMap()
{
  for (int j = 0; j < MAP_SIZE; j++)
  {
    for (int i = 0; i < MAP_SIZE; i++)
    {
        printf("%d ", input_map[j][i]);
    }
    putchar('\n');
  }
}

typedef struct YX { int y; int x; } YX;

void glassFire(int mask_Y, int mask_X)
{
  YX * stack = NULL;
  size_t sz = 0;
  size_t deep = 0;

  for (;;) {
    if (input_map[mask_Y][mask_X] == 1)
    {
      input_map[mask_Y][mask_X] = cnt;

      static const YX offset[] = {
        {0,1},  {1,0},  {1,1}, {1,-1}, {-1,1}, {-1,0}, {-1,-1},  {0,-1}
      };

      /* 8 positions to save */
      if ((deep + 8) > sz) {
        sz += 8;
        stack = realloc(stack, sz * sizeof(YX));
      }

      for (int dir = 0; dir != 8; ++dir) {
        int ny = mask_Y + offset[dir].y;
        int nx = mask_X + offset[dir].x;

        if ((ny >= 0) && (nx >= 0) &&
            (ny < MAP_SIZE) && (nx < MAP_SIZE) &&
            (input_map[ny][nx] == 1))
        {
          stack[deep].y = ny;
          stack[deep].x = nx;
          deep += 1;
        }
      }
    }

    if (deep == 0)
      /* empty */
      break;

    deep -= 1;
    mask_Y = stack[deep].y;
    mask_X = stack[deep].x;
  }

  free(stack);
}

void Serching() /* probably searching */
{
  PrintMap();

  for (int i = 0; i < MAP_SIZE; i++)
  {
    for (int j = 0; j < MAP_SIZE; j++)
    {
        if (input_map[j][i] == 1)
        {
          cnt++;
          glassFire(j, i);}
      }
   }
}

int main()
{
  Serching();
  putchar('\n');
  PrintMap();
}

如果我使用该完整程序（我删除了无用的 INILBCN ）：

pi@raspberrypi:~ $ gcc -pedantic -Wall -Wextra c.c
pi@raspberrypi:~ $ ./a.out
0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 
0 1 1 1 1 1 1 0 0 0 0 0 1 1 0 
0 1 0 0 0 0 1 0 0 0 1 1 0 1 0 
0 1 0 1 1 0 1 0 1 1 1 0 0 1 0 
0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 
0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 
0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 
0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 
0 1 0 0 0 0 0 0 1 0 1 0 0 1 0 
0 1 0 0 0 0 0 1 1 0 1 0 0 1 0 
0 1 1 1 1 1 0 1 0 0 1 1 0 1 0 
0 0 0 0 0 1 0 1 0 1 0 0 0 1 0 
0 1 1 1 0 1 0 1 0 0 0 0 0 1 0 
0 1 0 0 0 1 0 1 1 1 1 1 1 1 0 
0 1 1 1 1 1 0 0 0 0 0 0 0 1 0 

0 0 0 0 0 0 0 0 0 0 0 0 4 4 0 
0 2 2 2 2 2 2 0 0 0 0 0 4 4 0 
0 2 0 0 0 0 2 0 0 0 4 4 0 4 0 
0 2 0 2 2 0 2 0 4 4 4 0 0 4 0 
0 2 0 0 2 0 2 0 4 0 0 0 0 4 0 
0 2 2 0 2 0 2 0 4 0 0 5 0 4 0 
0 0 0 2 2 0 2 0 4 0 5 5 0 4 0 
0 3 0 2 2 0 2 0 4 0 5 0 0 4 0 
0 3 0 0 0 0 0 0 4 0 5 0 0 4 0 
0 3 0 0 0 0 0 4 4 0 5 0 0 4 0 
0 3 3 3 3 3 0 4 0 0 5 5 0 4 0 
0 0 0 0 0 3 0 4 0 5 0 0 0 4 0 
0 3 3 3 0 3 0 4 0 0 0 0 0 4 0 
0 3 0 0 0 3 0 4 4 4 4 4 4 4 0 
0 3 3 3 3 3 0 0 0 0 0 0 0 4 0

编译和执行：

cnt

我将input_map和{{1}}设为全局，但是它们可以位于 Serching 中，并可以为 glassFire 提供参数。全局变量越少越好。

请注意，我的代码产生的结果与您的代码（对校正结果取模）相同，但我不知道您的代码是否正确，因为我不知道它的真正作用是什么;-)

使用堆栈

1 个答案: