看数据结构写代码（13）栈的应用（四） 迷宫求解

这是一个 用 穷举法 解 迷宫问题 的一个示例，但在 效率 和 解的 最短路径上 就稍显不足了。

这 两个问题，留在 以后 空闲 时刻 解答。

欢迎指出代码不足

下面上代码：

// Maze.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <stdlib.h>struct Postion{int x;int y;};typedef int Direction;struct ElementType{Postion pos;Direction direction; };enum E_State{E_State_Error = 0,E_State_Ok = 1,};//链表节点nodestruct lStackNode{ElementType data;lStackNode * next;};//链栈struct linkStack{lStackNode * bottom;lStackNode * top;int len;};lStackNode * makeNode(ElementType data){lStackNode * pNode = (lStackNode *)malloc(sizeof(lStackNode));if (pNode != NULL){pNode->data = data;pNode->next = NULL;}return pNode;}E_State stackInit(linkStack * lStack){//分配头节点lStackNode * pNode = (lStackNode *) malloc(sizeof(lStackNode));if (pNode == NULL){return E_State_Error;}pNode->next = NULL;//栈顶和栈底指向同一个节点时为空.lStack->bottom = lStack->top = pNode;lStack->len = 0;return E_State_Ok;}void stackClear(linkStack * stack){lStackNode * next = stack->bottom->next;while (next != NULL){lStackNode * freeNode = next;/*又粗心了。。。free(freeNode);next = next->next;*/next = next->next;free(freeNode);}stack->top = stack->bottom;//忘记 了//错误stack->bottom->next = NULL;stack->len = 0;}void stackDestory(linkStack * stack){stackClear(stack);free(stack->top);stack->top = stack->bottom = NULL;}E_State stackGetTop(linkStack stack,ElementType * topData){//链表的栈顶 指向 栈顶元素if (stack.top != stack.bottom){*topData = stack.top->data;return E_State_Ok;}else{return E_State_Error;}}int stackLen(linkStack stack){return stack.len;}bool stackEmpty(linkStack stack){return stack.top == stack.bottom ? true : false;}E_State stackPush(linkStack * stack,ElementType data){lStackNode * node = makeNode(data);if (node != NULL){stack->top->next = node;stack->top = node;stack->len++;}else{return E_State_Error;}}E_State stackPop(linkStack * stack,ElementType * data){if (stack->top != stack->bottom){//首先指向第一个元素.lStackNode * next = stack->bottom;*data = stack->top->data;//找到栈顶元素的前驱while (next->next != stack->top){next = next->next;}free(stack->top);next->next = NULL;stack->top = next;//忘记加了stack->len--;return E_State_Ok;}else{return E_State_Error;}}int mazeData[10][10] = {{0,0,0,0,0,0,0,0,0,0},{0,1,1,0,1,1,1,0,1,0},{0,1,1,0,1,1,1,0,1,0},{0,1,1,1,1,0,0,1,1,0},{0,1,0,0,0,1,1,1,1,0},{0,1,1,1,0,1,1,1,1,0},{0,1,0,1,1,1,0,1,1,0},{0,1,0,0,0,1,0,0,1,0},{0,0,1,1,1,1,1,1,1,0},{0,0,0,0,0,0,0,0,0,0},};//pos坐标的节点 是否 可通过bool isPass(Postion pos){int pass = mazeData[pos.x][pos.y];//节点不是障碍物，并且 是没有通过的节点return pass == 1 ? true : false;}//获取下一个坐标Postion getNextPos(Postion pos,Direction di){int addArray[][2] = {{0,1},{1,0},{0,-1},{-1,0}};Postion nextPos;nextPos.x = pos.x + addArray[di][0];nextPos.y = pos.y + addArray[di][1];return nextPos;}void footPos(Postion pos,int curStep){mazeData[pos.x][pos.y] = curStep;}void delPos(Postion pos){mazeData[pos.x][pos.y] = -1;}int mazePath(Postion startPos,Postion endPos){linkStack stack;ElementType data;int curStep = 2;stackInit(&stack);do{if (isPass(startPos)){footPos(startPos,curStep);curStep++;data.pos.x = startPos.x;data.pos.y = startPos.y;data.direction = 0;stackPush(&stack,data);if (startPos.x == endPos.x && startPos.y == endPos.y){stackDestory(&stack);return 1;}else{startPos = getNextPos(startPos,0);}}else{if (!stackEmpty(stack)){stackPop(&stack,&data);curStep--;while (data.direction == 3 && !stackEmpty(stack)){//错误忘记 删除了..delPos(data.pos);stackPop(&stack,&data);curStep --;//startPos.x = data.pos.x;//startPos.y = data.pos.y;}if (data.direction < 3){data.direction++; // 换下一个方向探索stackPush(&stack,data);curStep++;startPos = getNextPos(data.pos,data.direction);}}}} while (!stackEmpty(stack));stackDestory(&stack);return 0;}// 输出解void PrintMaze(){ int i,j;int x = 10,y = 10;for(i=0;i<x;i++){for(j=0;j<y;j++)printf("%3d",mazeData[i][j]);printf("\n");}printf("\n");}int _tmain(int argc, _TCHAR* argv[]){printf("初始迷宫如下：\n");PrintMaze();Postion startPos;Postion endPos;printf("请输入起始位置 坐标x,y （从0行,0列算起）:");scanf("%d%d",&startPos.x,&startPos.y);printf("请输入终止位置 坐标x,y （从0行,0列算起）:");scanf("%d%d",&endPos.x,&endPos.y);int result = mazePath(startPos,endPos);if (result == 1){printf("迷宫解如下：（起始地址为2)\n");PrintMaze();}else{printf("迷宫从 (%d,%d) 至 (%d,%d) 无解\n",startPos.x,startPos.y,endPos.x,endPos.y);PrintMaze();}return 0;}

初始迷宫 中 0代表 墙，1 代表 通路。