指針二進制樹迷宮解算器在C
[英]Pointers Binary Tree Maze Solver in C
問題描述
我需要創建一個用C語言編寫的機器人模擬器。機器人必須使用遞歸回溯算法找到2d labirinth的出口，我知道這個算法是如何工作的，但我不知道如何實現它。 我想我可以使用指針使用二叉樹，但我不知道如何做到這一點，你能嘗試向我解釋一下嗎？
這是我創建的程序，現在機器人正在進入一個循環，因為改變方向的方法
#ifdef __unix__
#include <unistd.h>
#elif defined _WIN32
#include <windows.h>
#define sleep(x) Sleep(1000 * x)
#endif
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
void goUp();
void goDown();
void goLeft();
void goRight();
typedef struct robot {
int direction;
bool is_moving;
}robot;
typedef struct room {
robot robot;
bool is_robot;
int obstacle;
}room;
room Room[20][20];
int r = 12;
int c = 10;
void generation(room matrix[20][20])
{
srand(time(NULL));
int x,i,j;
x=0;
for(i=0;i<20;i++)
{
for(j=0;j<20;j++)
{
matrix[i][j].is_robot=false;
x=rand()%100+1;
if(x==1||x==50||x==100)
{
matrix[i][j].obstacle=1;
}
else
{
matrix[i][j].obstacle=0;
}
}
}
}
void print_matrix(room matrix[20][20])
{
int i,j;
for(i=0;i<20;i++)
{
for(j=0;j<20;j++)
{
if(matrix[i][j].obstacle==0)
{
if(matrix[i][j].is_robot==true)
{
printf("I");
}
else
{
printf(" ");
}
}
else
{
if(matrix[i][j].is_robot==true)
{
printf("I");
}
else
{
printf("o");
}
}
}
printf("\n");
}
}
bool changeDirection(room Room[20][20],int i,int j)
{
if(Room[i][j].robot.direction == 1)
{
if(Room[i-1][j].obstacle == 1 || i-1 == 0)
{
if(Room[i+1][j].obstacle == 1 || i+1 == 19)
{
Room[i][j].robot.direction = 2;
return true;
}
else
{
Room[i][j].robot.direction = 4;
return true;
}
}
else
{
Room[i][j].robot.direction = 3;
return true;
}
}
if(Room[i][j].robot.direction == 2)
{
if(Room[i-1][j].obstacle == 1 || i-1 == 0)
{
if(Room[i+1][j].obstacle == 1 || i+1 == 19)
{
Room[i][j].robot.direction = 1;
return true;
}
else
{
Room[i][j].robot.direction = 4;
return true;
}
}
else
{
Room[i][j].robot.direction = 3;
return true;
}
}
if(Room[i][j].robot.direction == 3)
{
if(Room[i][j+1].obstacle == 1 || j+1 == 19)
{
if(Room[i][j-1].obstacle == 1 || j-1 == 0)
{
Room[i][j].robot.direction = 4;
return true;
}
else
{
Room[i][j].robot.direction = 2;
return true;
}
}
else
{
Room[i][j].robot.direction = 1;
return true;
}
}
if(Room[i][j].robot.direction == 4)
{
if(Room[i][j+1].obstacle == 1 || j+1 == 19)
{
if(Room[i][j-1].obstacle == 1 || j-1 == 0)
{
Room[i][j].robot.direction = 3;
return true;
}
else
{
Room[i][j].robot.direction = 2;
return true;
}
}
else
{
Room[i][j].robot.direction = 1;
return true;
}
}
}
void goRight()
{
c=c+1;
Room[r][c].robot.direction=1;
Room[r][c].is_robot=true;
Room[r][c-1].is_robot=false;
}
void goLeft()
{
c=c-1;
Room[r][c].robot.direction=2;
Room[r][c].is_robot=true;
Room[r][c+1].is_robot=false;
}
void goUp()
{
r=r-1;
Room[r][c].robot.direction=3;
Room[r][c].is_robot=true;
Room[r+1][c].is_robot=false;
}
void goDown()
{
r=r+1;
Room[r][c].robot.direction=4;
Room[r][c].is_robot=true;
Room[r-1][c].is_robot=false;
}
int main()
{
generation(Room);
Room[r][c].robot.direction = 1;
Room[r][c].robot.is_moving = true;
Room[r][c].is_robot = true;
do
{
Room[r][c].robot.is_moving = true;
if (Room[r][c].robot.direction == 1 && Room[r][c].robot.is_moving == true) // destra
{
if(Room[r][c +1].obstacle == 1 || c+1 == 19)
{
changeDirection(Room,r,c);
}
else
{
goRight();
}
}
if (Room[r][c].robot.direction == 2 && Room[r][c].robot.is_moving == true) // sinistra
{
if(Room[r][c -1].obstacle == 1 || c-1 == 0)
{
changeDirection(Room,r,c);
}
else
{
goLeft();
}
}
if (Room[r][c].robot.direction == 3 && Room[r][c].robot.is_moving == true) // su
{
if(Room[r-1][c].obstacle == 1 || r-1 == 0)
{
changeDirection(Room,r,c);
}
else
{
goUp();
}
}
if (Room[r][c].robot.direction == 4 && Room[r][c].robot.is_moving == true) // giu
{
if(Room[r+1][c].obstacle == 1 || r+1 == 19)
{
changeDirection(Room,r,c);
}
else
{
goDown();
}
}
print_matrix(Room);
sleep(0.1);
system("cls");
}
while(1);
print_matrix(Room);
}
1 個解決方案
解決方案1
0 2016-10-22 13:45:17
我很難理解二叉樹如何在迷宮中找到一條路徑（也許它用來代表迷宮？）但也許我是盲目的。 我只是制作一個2d int數組，讓0意味着位置被阻擋（那里有一堵牆或其他東西）和1意味着它是開放的（你可以移動那里）。 蠻力回溯程序，正常運動（左，右，上，下）將是：
f(x,y){
// you found the place your want to go to
if (x,y) is (destinationX,destinationY)
return true
block the position (x,y) // i.e. mark current position as visited
if there is an open spot at (x,y-1) AND f(x,y-1)
return true
if there is an open spot at (x,y+1) AND f(x,y+1)
return true
if there is an open spot at (x-1,y) AND f(x-1,y)
return true
if there is an open spot at (x+1,y) AND f(x+1,y)
return true
return false
}
假設你的迷宮看起來像：
"+" is where you start ([1][1])
"-" is your destination ([3][1])
"#" is a blocked region
===========
|#|#|#|#|#|
|#|+| |#|#|
|#|#| |#|#|
|#|-| | |#|
|#|#|#|#|#|
===========
使用上述想法我有：
#include <stdio.h>
#define width 5
#define height 5
// print maze
void print(char arr[][width]){
for (int i = 0; i < 2*width+1; i++) printf("=");
printf("\n");
for (int i = 0; i < height; i++) {
for (int j = 0; j < width; j++) {
printf("|%c",arr[i][j]);
}
printf("|\n");
}
for (int i = 0; i < 2*width+1; i++) printf("=");
}
// starting from (x,y) to (destX,destY)
int path(int arr[][width],int x,int y,int destX,int destY,char toDest[][width]){
if (x==destX && y==destY) {
toDest[y][x] = '*';
print(toDest);
return 1;
}
// mark current position as visited
arr[y][x] = 0;
toDest[y][x] = '*';
// left
if (arr[y][x-1] && path(arr,x-1,y,destX,destY,toDest))
return 1;
// right
if (arr[y][x+1] && path(arr,x+1,y,destX,destY,toDest))
return 1;
// up
if (arr[y-1][x] && path(arr,x,y-1,destX,destY,toDest))
return 1;
// down
if (arr[y+1][x] && path(arr,x,y+1,destX,destY,toDest))
return 1;
return 0;
}
int main () {
// use this to store path
// and then print it out if found
char toDest[height][width] = {
{'#','#','#','#','#'},
{'#',' ',' ','#','#'},
{'#','#',' ','#','#'},
{'#',' ',' ',' ','#'},
{'#','#','#','#','#'}
};
// 0 -> position is blocked
// 1 -> position is open
int maze[height][width] = {
{0,0,0,0,0},
{0,1,1,0,0},
{0,0,1,0,0},
{0,1,1,1,0},
{0,0,0,0,0}
};
path(maze,1,1,1,3,toDest);
}
輸出：
===========
|#|#|#|#|#|
|#|*|*|#|#|
|#|#|*|#|#|
|#|*|*| |#|
|#|#|#|#|#|
===========
在輸出中，路徑由* s指定
在C中回溯迷宮求解器
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