POJ 2251 Dungeon Master (BFS)

迷宫逃脱算法解析
Dungeon Master

Time Limit: 1000MS
Memory Limit:65536K

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


还是vis判断好用!

完整代码:
/*0ms,684KB*/

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int dx[6] = {1, -1, 0, 0, 0, 0};
const int dy[6] = {0, 0, 1, -1, 0, 0};
const int dz[6] = {0, 0, 0, 0, 1, -1};
const int maxn = 35;

struct node
{
	int x, y, z, step;
	node() {}
	node(int a, int b, int c, int d)
	{
		x = a, y = b, z = c, step = d;
	}
} tmp;
queue<node> q;
int n, m, t, x, y, z, i, j, k;
bool vis[maxn][maxn][maxn];
char mp[maxn][maxn][maxn];

bool bfs()
{
	while (!q.empty())
	{
		tmp = q.front();
		q.pop();
		if (mp[tmp.z][tmp.y][tmp.x] == 'E')
		{
			printf("Escaped in %d minute(s).\n", tmp.step);
			return true;
		}
		for (i = 0; i < 6; ++i)
		{
			x = tmp.x + dx[i], y = tmp.y + dy[i], z = tmp.z + dz[i];
			if (mp[z][y][x] != '#' && !vis[z][y][x])
			{
				vis[z][y][x] = true;
				q.push(node(x, y, z, tmp.step + 1));
			}
		}
	}
	return false;
}

int main()
{
	while (~scanf("%d%d%d\n", &t, &n, &m), t)
	{
		memset(vis, 0, sizeof(vis));
		memset(mp, '#', sizeof(mp));
		while (!q.empty()) q.pop();
		for (i = 1; i <= t; ++i)
		{
			for (j = 1; j <= n; ++j)
			{
				for (k = 1; k <= m; ++k)
				{
					mp[i][j][k] = getchar();
					if (mp[i][j][k] == 'S')
					{
						vis[i][j][k] = true;
						q.push(node(k, j, i, 0));
					}
				}
				getchar();
			}
			getchar();
		}
		if (!bfs()) puts("Trapped!");
	}
	return 0;
}

【无人机】基于改进粒子群算法的无人机路径规划研究[和遗传算法、粒子群算法进行比较](Matlab代码实现)内容概要:本文围绕基于改进粒子群算法的无人机路径规划展开研究,重点探讨了在复杂环境中利用改进粒子群算法(PSO)实现无人机三维路径规划的方法,并将其与遗传算法(GA)、标准粒子群算法等传统优化算法进行对比分析。研究内容涵盖路径规划的多目标优化、避障策略、航路点约束以及算法收敛性和寻优能力的评估,所有实验均通过Matlab代码实现,提供了完整的仿真验证流程。文章还提到了多种智能优化算法在无人机路径规划中的应用比较,突出了改进PSO在收敛速度和全局寻优方面的优势。; 适合人群:具备一定Matlab编程基础和优化算法知识的研究生、科研人员及从事无人机路径规划、智能优化算法研究的相关技术人员。; 使用场景及目标:①用于无人机在复杂地形或动态环境下的三维路径规划仿真研究;②比较不同智能优化算法(如PSO、GA、蚁群算法、RRT等)在路径规划中的性能差异;③为多目标优化问题提供算法选型和改进思路。; 阅读建议:建议读者结合文中提供的Matlab代码进行实践操作,重点关注算法的参数设置、适应度函数设计及路径约束处理方式,同时可参考文中提到的多种算法对比思路,拓展到其他智能优化算法的研究与改进中。
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