迷宫问题。easy的。用递归和非递归。
/*
* 问题描述:
* 说明老鼠走迷宫是递回求解的基本题型，我们在二维阵列中使用1表示迷宫墙壁，使用2来表
* 示老鼠的行走路径，试以程式求出由入口至出口的路径。
*
* 若更改迷宫数据，需要修改行数：MAP_Y，列数：MAP_X即可.另外迷宫的最外层都必须设置为墙壁，
* 起始点和终点定位可修改。
*/
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <stack>
using namespace std;
#define MAP_X 11
#define MAP_Y 12
#define WALL 1
#define PASS 0
#define FOOT 2
#define MARK 3
/*
* 坐标系统：
* +----------------->(X)
* |
* | (x,y)
* |
* |
* |
* (Y)
*/
typedef struct _pos{
int x;
int y;
}pos;
//起始位置：第1行1列
static int start[2]= {1,1};
//结束位置：第MAX_Y-2行MAP_X-2列
static int end[2]= {MAP_Y-2,MAP_X-2};
static int finish=0;
/*
* 迷宫最外一层为边界墙壁，从[1,1]起点到[MAP_X,MAP_Y]终点
*
*/
/*
测试迷宫
设定MAP_X=7，MAX_Y=7即可
int m[MAP_X][MAP_Y] =
{
{WALL, WALL, WALL, WALL, WALL, WALL, WALL},
{WALL, PASS, PASS, PASS, PASS, PASS, WALL},
{WALL, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, PASS, WALL, PASS, WALL, WALL},
{WALL, WALL, PASS, WALL, PASS, WALL, WALL},
{WALL, PASS, PASS, PASS, PASS, PASS, WALL},
{WALL, WALL, WALL, WALL, WALL, WALL, WALL}
};
*/
/*
*行数(MAP_Y)12*列数(MAP_X)11
*成功迷宫
*/
int m[MAP_Y][MAP_X] = {
{WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL},/*一行*/
{WALL, PASS, PASS, PASS, PASS, PASS, WALL, WALL, WALL, WALL, WALL},
{WALL, WALL, WALL, WALL, WALL, PASS, WALL, PASS, WALL, WALL, WALL},
{WALL, WALL, WALL, WALL, WALL, PASS, WALL, WALL, WALL, WALL, WALL},
{WALL, PASS, WALL, PASS, WALL, PASS, WALL, PASS, PASS, PASS, WALL},
{WALL, PASS, WALL, PASS, WALL, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, PASS, PASS, WALL, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, WALL, WALL, WALL, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, WALL, PASS, PASS, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, PASS, PASS, WALL, WALL, WALL, PASS, PASS, PASS, WALL},
{WALL, WALL, WALL, PASS, PASS, PASS, PASS, PASS, PASS, PASS, WALL},
{WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL}
};
/*
* 范例，失败迷宫
*/
int mg2[MAP_Y][MAP_X] = {
{WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL},/*一行*/
{WALL, PASS, PASS, PASS, PASS, PASS, WALL, WALL, WALL, WALL, WALL},
{WALL, WALL, WALL, WALL, WALL, PASS, WALL, PASS, WALL, WALL, WALL},
{WALL, WALL, WALL, WALL, WALL, PASS, WALL, WALL, WALL, WALL, WALL},
{WALL, PASS, WALL, PASS, WALL, PASS, WALL, PASS, PASS, PASS, WALL},
{WALL, PASS, WALL, PASS, WALL, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, PASS, PASS, WALL, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, WALL, WALL, WALL, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, WALL, PASS, PASS, PASS, WALL, PASS, WALL, PASS, WALL},
{WALL, PASS, PASS, PASS, WALL, WALL, WALL, PASS, PASS, WALL, WALL},
{WALL, WALL, WALL, PASS, PASS, PASS, PASS, PASS, WALL, PASS, WALL},
{WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL, WALL}
};
/*
* 打印迷宫
* w=11,h=12
*/
void display(int m[MAP_Y][MAP_X],int w,int h)
{
int i,j;
/*一层层打印*/
for(i=0; i<h; i++) {
/*一列列打印*/
for(j=0; j<w; j++) {
if(m[i][j]==WALL) {
/*draw wall*/
printf("■");//△
} else if(m[i][j]==FOOT) {
/*draw foot*/
printf("★");
} else if(m[i][j]==PASS) {
/*draw road*/
printf("□");
} else {
/*标记过的路径*/
printf("TT");
}
}
printf("\n");
}
}
/*
* 迷宫的试探，使用递归实现。
*
*/
int visit(int m[MAP_Y][MAP_X],int x,int y)
{
m[x][y]=FOOT;
/*当前位置到达end点，探路成功！*/
if( (x == end[0]) && (y == end[1]) ) {
finish=1;
}
/*往右走，若通路，向右一步；若不通，继续*/
if(!finish&&m[x][y+1]==PASS) {
printf("右{%d %d} ",x,y+1);
visit(m,x,y+1);
}
/*往左探路*/
if(!finish&&m[x][y-1]==PASS) {
printf("左{%d %d} ",x,y-1);
visit(m,x,y-1);
}
/*往下探路*/
if(!finish&&m[x+1][y]==PASS) {
printf("下{%d %d} ",x+1,y);
visit(m,x+1,y);
}
/*往上探路*/
if(!finish&&m[x-1][y]==PASS) {
printf("上{%d %d} ",x-1,y);
visit(m,x-1,y);
}
if(finish!=1) {
m[x][y]=PASS;
}
return finish;
}
/*
* 迷宫的试探，使用非递归实现。
*
*/
int visit2(int m[MAP_Y][MAP_X])
{
stack<pos> mystack;
/*当前位置设置*/
pos cur_pos;
cur_pos.x = start[0];
cur_pos.y = start[1];
do{
if ((cur_pos.x == end[1]) && (cur_pos.y == end[0]))
{
mystack.push(cur_pos);
break;
}
/*往右走，若通路，向右一步；若不通，继续*/
if(m[cur_pos.y][cur_pos.x+1]==PASS) {
//printf("右{%d %d} ",cur_pos.y,cur_pos.x+1);
/*当前点压栈，设置当前点为右边邻接点*/
m[cur_pos.y][cur_pos.x] = MARK;//标记为已走过
mystack.push(cur_pos);
cur_pos.x = cur_pos.x+1;
continue;
}
/*往左探路*/
if(m[cur_pos.y][cur_pos.x-1]==PASS) {
//printf("左{%d %d} ",cur_pos.y,cur_pos.x-1);
/*当前点压栈，设置当前点为左边邻接点*/
m[cur_pos.y][cur_pos.x] = MARK;//标记为已走过
mystack.push(cur_pos);
cur_pos.x = cur_pos.x-1;
continue;
}
/*往下探路*/
if(m[cur_pos.y+1][cur_pos.x]==PASS) {
//printf("下{%d %d} ",cur_pos.y+1,cur_pos.x);
/*当前点压栈，设置当前点为下边邻接点*/
m[cur_pos.y][cur_pos.x] = MARK;//标记为已走过
mystack.push(cur_pos);
cur_pos.y = cur_pos.y+1;
continue;
}
/*往上探路*/
if(m[cur_pos.y-1][cur_pos.x]==PASS) {
//printf("上{%d %d} ",cur_pos.y-1,cur_pos.x);
/*当前点压栈，设置当前点为上边邻接点*/
m[cur_pos.y][cur_pos.x] = MARK;//标记为已走过
mystack.push(cur_pos);
cur_pos.y = cur_pos.y-1;
continue;
}
/*周围无可走之路，回退*/
if(mystack.size() > 0){
m[cur_pos.y][cur_pos.x] = MARK;//标记为已走过
cur_pos = mystack.top();
mystack.pop();
//getchar();
continue;
}else{
/*无法回退，搜索失败*/
return -1;
}
}while (!mystack.empty());
if(mystack.empty()){
return -1;
}
/*探寻成功，设置路径*/
while (!mystack.empty())
{
cur_pos = mystack.top();
m[cur_pos.y][cur_pos.x] = FOOT;//标记为已走过
mystack.pop();
}
return 0;
}
int main()
{
/*使用递归方法*/
//if(visit(m,start[0],start[1])==0) {
/*使用堆栈方法*/
printf("--------------------地图1-------------------\n\n");
display(m,MAP_X,MAP_Y);
if(visit2(m) < 0){
printf("\n没通路!");
//display(m,MAP_X,MAP_Y);
} else {
printf("\n找到出口！\n\n");
display(m,MAP_X,MAP_Y);
}
printf("\n--------------------地图2-------------------\n\n");
display(mg2,MAP_X,MAP_Y);
if(visit2(mg2) < 0){
printf("\n没通路!");
//display(mg2,MAP_X,MAP_Y);
} else {
printf("\n找到出口！\n\n");
display(mg2,MAP_X,MAP_Y);
}
system("pause > NULL");
return 0;
}
