POJ-2251 Dungeon Master(BFS)

最新推荐文章于 2022-10-14 19:40:10 发布

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最新推荐文章于 2022-10-14 19:40:10 发布

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分类专栏：搜索文章标签： BFS ACM

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本文详细解析了一个三维迷宫逃脱问题的算法实现，利用BFS（宽度优先搜索）算法找到从起点到终点的最短路径。文章介绍了问题背景，给出了具体的输入输出样例，并提供了完整的C++代码实现。

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Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 50464		Accepted: 18983

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!

题意：

给你一个三维的矩阵，有开始点，有结束点，每次可以走东、西、南、北、上、下。

其中有些点为石头，不能通过，现在求从开始点走到结束点的最短时间。如果不能走到则输出"Trapped!"。

思路：

一开始用的DFS,超时了。改成BFS就好了，直接套模板即可。。。

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 32;

char maze[MAXN][MAXN][MAXN];
bool book[MAXN][MAXN][MAXN];
int x2, y2, z2;
int move[6][3] = { {0,0,1}, {0,1,0}, {0,-1,0}, {0,0,-1}, {1,0,0}, {-1,0,0}};
int Min, L, R, C;;
struct State
{
	int x, y, z;
	int Step;
};
bool check(State s)
{
	if(s.x >= L || s.x < 0 || s.y >= R || s.y < 0 || s.z >= C || s.z < 0)
		return 0;
	if(book[s.x][s.y][s.z] == false && maze[s.x][s.y][s.z] != '#') {
		return 1;
	}
	return 0;
}
void bfs(State s)
{
	queue <State> q;
	State now, next;
	q.push(s);
	book[s.x][s.y][s.z] = true;
	while(!q.empty())
	{
		now = q.front();
		if(now.x == x2 && now.y == y2 && now.z == z2){
			Min = now.Step;
			return ;
		}
		for(int i = 0; i < 6; i++) {
			next.x = now.x + move[i][0];
			next.y = now.y + move[i][1];
			next.z = now.z + move[i][2];
			if(check(next)) {
				next.Step = now.Step + 1;
				q.push(next);
				book[next.x][next.y][next.z] = true;
			}
		}
		q.pop();
	}
	return ;
	
}
int main()
{
	
	while(scanf("%d%d%d", &L, &R, &C)!=EOF)
	{
		if(L == 0 && R == 0 && C == 0)
			break;
		memset(book, false, sizeof(book));
		int x1, y1, z1;
		for(int i = 0; i < L; i++) 
			for(int j = 0; j < R; j++) {
				scanf("%s", maze[i][j]);
				for(int k = 0; k < C; k++) {
					if(maze[i][j][k] == 'S') {
						x1 = i;y1 = j; z1 = k;
					}
					if(maze[i][j][k] == 'E') {
						x2 = i; y2 = j; z2 = k;
					}
				} 
					
			}
				
		State s;
		s.x = x1; s.y = y1; s.z = z1; s.Step = 0;
			
		Min = 9999999;
		bfs(s);
		if(Min == 9999999)	
			printf("Trapped!\n");
		else
			printf("Escaped in %d minute(s).\n", Min);		
	
	} 
	
	return 0;
}