1449 - 3Ddungeon(3D迷宫）_控制台3d字符迷宫-优快云博客

本文介绍了一种解决3D迷宫问题的算法，通过使用三维数组和BFS（宽度优先搜索）算法，解决了在限定时间内从起点到达终点的问题。文章详细解释了如何通过遍历每个单元格来寻找最短路径，并提供了完整的代码实现。

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1449 - 3Ddungeon

时间限制：1秒内存限制：128兆

题目描述

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

Is an escape possible? If yes, how long will it take?

输入

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

输出

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form
Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line
Trapped!

样例输入

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

样例输出

Escaped in 11 minute(s).
Trapped!

其实和平面迷宫的走法是一样的，只是地图需要三维数组来标记而已，

走法也从上下左右变成了六个方向，由于地图大小最大为30*30*30，

所以此处使用bfs求解

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
int sx, sy, sz;
int ex, ey, ez;
char MAP[35][35][35];
int xx[6]={1, -1, 0, 0, 0, 0};
int yy[6]={0, 0, 1, -1, 0, 0};
int zz[6]={0, 0, 0, 0, 1, -1};
struct Node
{
	int x, y, z, steps;
}NOW,NEXT;
bool vis[35][35][35];
int bfs(int x1, int y1,int z1)
{
	memset(vis,0,sizeof(vis));
	NOW.x = sx; NOW.y = sy; NOW.z = sz ; NOW.steps = 0; vis[sx][sy][sz] = true;
	queue<Node> q;
	q.push(NOW);
	while(!q.empty())
	{
		NOW = q.front();
		if(NOW.x  == ex && NOW.y == ey && NOW.z == ez)	return NOW.steps;
		q.pop();
		for(int i = 0; i < 6; i++)
		{
			int X = NOW.x + xx[i];
			int Y = NOW.y + yy[i];
			int Z = NOW.z + zz[i];
			if(X >= 0 && Y >= 0 &&  Z >= 0 && X < x1 && Y < y1 && Z < z1 && vis[X][Y][Z] == false && MAP[X][Y][Z] != '#')
			{
				vis[X][Y][Z] = true;
				NEXT.x = X; NEXT.y = Y; NEXT.z = Z; NEXT.steps = NOW.steps + 1;
				q.push(NEXT);
			}
		}
	}
	return -1;
}
int main()
{
    int x1, y1, z1;
	while(scanf("%d %d %d", &x1, &y1, &z1))
	{
		getchar();
		if(!x1 || !y1 || !z1)
		return 0;
		for(int i=0; i<x1; i++)
		{
			for(int j=0; j<y1; j++)
			{
				gets(MAP[i][j]);
			}
			getchar();
		}
		for(int i=0; i<x1; i++)
		{
			for(int j=0; j<y1; j++)
			{
				for(int k=0; k<z1; k++)
				{
					if(MAP[i][j][k] == 'S')
					{
						sx = i;
						sy = j;
						sz = k;
					}
					if(MAP[i][j][k] == 'E')
					{
						ex = i;
						ey = j;
						ez = k;
					}
				}
			}
		}
		int min=bfs(x1, y1, z1);
		if(min!=-1)
			printf("Escaped in %d minute(s).\n",min);
		else
			printf("Trapped!\n");
	}
}