Dungeon Master （三维bfs）

最新推荐文章于 2022-03-28 21:47:12 发布

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最新推荐文章于 2022-03-28 21:47:12 发布

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文章标签： bfs

本文链接：https://blog.youkuaiyun.com/weixin_41988889/article/details/95787460

博客围绕被困3D地牢寻找最快逃脱路径展开。地牢由单位立方体组成，移动需一分钟，不能斜向移动。输入包含地牢层数、行列数等信息，输出为能否逃脱及最短逃脱时间。强调不用管输入格式，测试数据由系统自动给出。

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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?

Input

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.

Output

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!

Sample Input

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

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

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

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

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

注意输入的格式我们不用管！测试数据由系统自动给出！！！

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int dx[] = {-1,0,1,0,0,0};
int dy[] = {0,1,0,-1,0,0};
int dz[] = {0,0,0,0,1,-1};
int vis[35][35][35];
char maze[35][35][35];
struct node{
	int x,y,z;
	int step;
};
int L,R,C,ans;
void bfs(int x,int y,int z){
	queue<node> q;
	node now,next;
	now.x = x,now.y = y,now.z = z,now.step = 0;
	q.push(now);
	while(q.size()){
		now = q.front();
		q.pop();
		for(int i = 0;i < 6;i ++){
			next.x =now.x + dx[i],next.y = now.y + dy[i],next.z = now.z + dz[i],next.step = now.step + 1;
			if(next.x >= 0 && next.x < L && next.y >= 0 && next.y < R && next.z >= 0 && next.z < C && vis[next.x][next.y][next.z]){
				vis[next.x][next.y][next.z] = 0;
				q.push(next);
				if(maze[next.x][next.y][next.z] == 'E'){
					ans = next.step;
					return;
				}
			}
		}
	}
}
int main()
{
	while(cin>>L>>R>>C&&L&&R&&C){
		ans = 0;
		int x,y,z;
		memset(vis,0,sizeof(vis));
		for(int i = 0;i < L;i ++){
			for(int j = 0;j < R;j ++){
				for(int k = 0;k < C;k ++){
					cin>>maze[i][j][k];
					if(maze[i][j][k] == 'S'){
						x = i,y = j,z = k;
					}
					if(maze[i][j][k] == '.' || maze[i][j][k] == 'E' ){
						vis[i][j][k] = 1;
					}
				}
			}
             // 刚开始在这里加了 cout<<endl;  一直报错！！注意输入的格式我们不用管！测试数据由系统自动给出！
		}
		bfs(x,y,z);
		if(ans == 0) cout<<"Trapped!"<<endl;
		else printf("Escaped in %d minute(s).\n",ans);
	}
	return 0;	
}