POJ-2251 Dungeon Master （搜索）

最新推荐文章于 2022-11-02 20:38:04 发布

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最新推荐文章于 2022-11-02 20:38:04 发布

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分类专栏：简单搜索

本文链接：https://blog.youkuaiyun.com/chenshibo17/article/details/80948052

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本文介绍了一个基于三维空间的迷宫逃脱问题，并通过BFS（宽度优先搜索）算法来寻找从起点S到终点E的最短路径。该算法在限定的时间内找到最快的逃脱路线，或判断无法逃脱。

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46122		Accepted: 17400

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

其实就是一个简单的三维空间的BFS搜索，从S到E不能经过#，可上，下，左，右，前，后，之类的，水题；

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
char m[40][40][40];
bool vis[40][40][40];
int a, b, c;
int fx[6][3] = { 1,0,0,0,1,0,0,0,1,-1,0,0,0,-1,0,0,0,-1 };
int ans = -1;
struct fuck
{
	int x, y, z;
	int step;
};
void BFS(int x,int y,int z)
{
	vis[x][y][z] = 1;
	queue<fuck>q;
	fuck st;
	st.x = x;st.y = y;st.z = z;st.step = 0;
	q.push(st);
	while (!q.empty())
	{
		fuck ste = q.front();
		q.pop();
		if (m[ste.x][ste.y][ste.z] == 'E')
		{
			ans = ste.step;
			while (!q.empty())
			{
				q.pop();
			}
			return;
		}
		for (int s = 0; s < 6; s++)
		{
			int xx = ste.x + fx[s][0];
			int yy = ste.y + fx[s][1];
			int zz = ste.z + fx[s][2];
			if (xx >= a || xx<0 || yy>=b || yy<0 || zz>=c || zz < 0||m[xx][yy][zz]=='#'||vis[xx][yy][zz])
			{
				continue;
			}
			fuck hh;
			hh.x = xx, hh.y = yy, hh.z = zz, hh.step = ste.step + 1;
			vis[xx][yy][zz] = 1;
			q.push(hh);
		}
	}
}
int main()
{
	while (cin >> a >> b >> c&&a&&b&&c)
	{
		ans = -1;
		int x, y, z;
		memset(vis, 0, sizeof(vis));
		for (int s = 0; s<a; s++)
		{
			for (int w = 0; w<b; w++)
			{
				for (int e = 0; e<c; e++)
				{
					cin >> m[s][w][e];
					if (m[s][w][e] == 'S')
					{
						x = s, y = w, z = e;
					}
				}
			}
		}
		BFS(x, y, z);
		if (ans == -1)
		{
			cout << "Trapped!" << endl;
		}
		else
		{
			cout << "Escaped in " << ans << " minute(s)." << endl;
		}
	}
	return 0;
}