C语言迷宫求解算法实现 - 使用链栈数据结构

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

#define row 8 #define col 11

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

typedef struct { int top; int right; int left; int bottom; } Direction;

typedef struct StackNode { Position position; Direction direct; struct StackNode* next; } StackNode, *LinkStackPtr;

typedef struct { LinkStackPtr top; int count; } LinkStack;

typedef struct { int x; int y; int seq; } Path;

int map[row][col] = { {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1}, {1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1}, {1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1}, {1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1}, {1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} }; // 迷宫地图，1代表墙壁，0代表路

int enterx = 1, entery = 1; // 入口坐标 int exitx = 6, exity = 9; // 出口坐标

Path path[(row - 1) * (col - 1)]; // 探索迷宫的路径 int pathLength = 0; // 路径长度

bool Panduan(int x, int y); bool Maze(LinkStack* L); bool IsEmpty(LinkStack *L); void CreateLink(LinkStack *L); LinkStackPtr GetTop(LinkStack *L); void Push(LinkStack *L, LinkStackPtr s); void Pop(LinkStack *L); void RecordExplorationPath(LinkStackPtr m); void PrintPath();

int main(int argc, char* argv[]) { LinkStack L; L.count = 0; CreateLink(&L);

bool flag = Maze(&L);
	
if (flag == true) {
	PrintPath();
} else {
	printf("Can not find path");
}

return 0;

}

// 判断是否是走过的迷宫块 bool Panduan(int x, int y) { if (map[x][y] == 0) { return true; }

return false;

}

// 探索迷宫可通行的路径 bool Maze(LinkStack* L) { bool flag = false;

LinkStackPtr p;
p = (LinkStackPtr)malloc(sizeof(struct StackNode));
p->position.x = enterx;
p->position.y = entery;
p->direct.top = 1;
p->direct.right = 0;
p->direct.left = 0;
p->direct.bottom = 0;
map[p->position.x][p->position.y] = 2;

// 将入口结点入栈
Push(L, p);

while (!IsEmpty(L)) {
	LinkStackPtr top = GetTop(L);
	
	if (top->position.x == exitx && top->position.y == exity) {
		flag = true;
		break;
	}
	
	bool moved = false;
	
	if (top->direct.top && Panduan(top->position.x - 1, top->position.y)) {
		top->direct.top = 0;
		Push(L, &(StackNode) { .position = { top->position.x - 1, top->position.y }, .direct = { 1, 0, 0, 0 }, .next = NULL });
		map[top->position.x - 1][top->position.y] = 2;
		RecordExplorationPath(GetTop(L));
		moved = true;
	} else if (top->direct.right && Panduan(top->position.x, top->position.y + 1)) {
		top->direct.right = 0;
		Push(L, &(StackNode) { .position = { top->position.x, top->position.y + 1 }, .direct = { 0, 1, 0, 0 }, .next = NULL });
		map[top->position.x][top->position.y + 1] = 2;
		RecordExplorationPath(GetTop(L));
		moved = true;
	} else if (top->direct.left && Panduan(top->position.x, top->position.y - 1)) {
		top->direct.left = 0;
		Push(L, &(StackNode) { .position = { top->position.x, top->position.y - 1 }, .direct = { 0, 0, 1, 0 }, .next = NULL });
		map[top->position.x][top->position.y - 1] = 2;
		RecordExplorationPath(GetTop(L));
		moved = true;
	} else if (top->direct.bottom && Panduan(top->position.x + 1, top->position.y)) {
		top->direct.bottom = 0;
		Push(L, &(StackNode) { .position = { top->position.x + 1, top->position.y }, .direct = { 0, 0, 0, 1 }, .next = NULL });
		map[top->position.x + 1][top->position.y] = 2;
		RecordExplorationPath(GetTop(L));
		moved = true;
	}
	
	if (!moved) {
		Pop(L);
	}
}

return flag;

}

// 判断链栈是否为空 bool IsEmpty(LinkStack *L) { if (L->count == 0) return true; return false; }

// 创建链栈 void CreateLink(LinkStack *L) { L->top = (LinkStackPtr)malloc(sizeof(struct StackNode)); if (!L->top) { exit(1); } L->top = NULL; L->count = 0; }

// 获取栈顶 LinkStackPtr GetTop(LinkStack *L) { return L->top; }

// 进栈 void Push(LinkStack *L, LinkStackPtr s) { s->next = L->top; L->top = s; L->count++; }

// 出栈 void Pop(LinkStack *L) { LinkStackPtr p; if (IsEmpty(L)) { return; } p = L->top; L->top = L->top->next; free(p); L->count--; }

// 记录探索路径 void RecordExplorationPath(LinkStackPtr m){ path[pathLength].x = m->position.x; path[pathLength].y = m->position.y; path[pathLength].seq = pathLength + 1; pathLength += 1; }

// 打印迷宫可通行的路径 void PrintPath(){ for (int i = 0; i < pathLength; i++) { printf("%d:(%d, %d)\n",path[i].seq, path[i].x, path[i].y); } return;