【POJ】2251 - Dungeon Master（dfs + 队列）

FishWang

发布于 2025-08-26 19:57:35

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代码可运行

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Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26403		Accepted: 10271

Description

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!

Source

Ulm Local 1997

还是bfs加队列。

代码如下：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
char mapp[35][35][35];
int a,b,c;		//地图总大小 
int st_x,st_y,st_z,endd_x,endd_y,endd_z;
int move_x[6] = {0,0,0,0,1,-1};
int move_y[6] = {0,0,1,-1,0,0};
int move_z[6] = {1,-1,0,0,0,0};
bool used[35][35][35];
struct node
{
	int x,y,z;
	int step;
}pr,ne;

bool check(node t)
{
	if (t.x < 1 || t.x > a || t.y < 1 || t.y > b || t.z < 1 || t.z > c || used[t.x][t.y][t.z] || mapp[t.x][t.y][t.z] == '#')
		return false;		//不能移动
	return true; 
}
int bfs()
{
	queue<node> q;
	CLR(used,0);
	pr.x = st_x;
	pr.y = st_y;
	pr.z = st_z;
	pr.step = 0;
	q.push(pr);
	used[pr.x][pr.y][pr.z] = true;
	while (!q.empty())
	{
		pr = q.front();
		q.pop();
		for (int i = 0 ; i < 6 ; i++)
		{
			ne = pr;
			ne.x += move_x[i];
			ne.y += move_y[i];
			ne.z += move_z[i];
			ne.step++;
			if (!check(ne))
				continue;
			if (ne.x == endd_x && ne.y == endd_y && ne.z == endd_z) 
				return ne.step;
			used[ne.x][ne.y][ne.z] = true;		//在这里标记而不是外面取数的时候标记 
			q.push(ne);
		}
	}
	return -1;
}
int main()
{
	while (~scanf ("%d %d %d",&a,&b,&c) && (a || b || c))
	{
		for (int i = 1 ; i <= a ; i++)
		{
			for (int j = 1 ; j <= b ; j++)
			{
				scanf ("%s",mapp[i][j]+1);
				for (int k = 1 ; k <= c ; k++)
				{
					if (mapp[i][j][k] == 'S')
					{
						st_x = i;
						st_y = j;
						st_z = k;
					}
					else if (mapp[i][j][k] == 'E')
					{
						endd_x = i;
						endd_y = j;
						endd_z = k;
					}
				}
			}
		}
		int ans = bfs();
		if (ans == -1)
			printf ("Trapped!\n");
		else
			printf ("Escaped in %d minute(s).\n",ans);
	}
	return 0;
}

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原始发表：2025-08-26，如有侵权请联系 cloudcommunity@tencent.com 删除

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