POJ2251 Dungeon Master(bfs)

YOLOv8使用设备摄像头实时监测 阅读详情

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!



三维迷宫,用一个三维数组保存然后跟二维一样的板子。S是起点E是终点,通过.可以去另一张图,能到达E输出步数,不能到达E则

输出Trapped!


AC代码:


#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;

typedef class
{
public:
	int l, r, c;
	int depth;
}SE;
SE s, e;
bool maze[40][40][40];
int ans;

bool bfs(int k, int i, int j)
{
	bool vis[40][40][40] = {false};
	SE queue[30000];
	int head, tail;
	queue[head = 0].l = k;
	queue[tail = 0].r = i;
	queue[0].c = j;
	queue[tail++].depth = 0;
	vis[k][i][j] = true;
	while(head < tail) {
		SE x = queue[head++];
		if(x.l == e.l && x.r == e.r && x.c == e.c) {
			ans = x.depth;
			return true;
		}
		if(maze[x.l][x.r][x.c - 1] && !vis[x.l][x.r][x.c - 1]) {
			vis[x.l][x.r][x.c - 1] = true;
			queue[tail].l = x.l;
			queue[tail].r = x.r;
			queue[tail].c = x.c - 1;
			queue[tail++].depth = x.depth + 1;
		}
		if(maze[x.l][x.r - 1][x.c] && !vis[x.l][x.r - 1][x.c]) {
			vis[x.l][x.r - 1][x.c] = true;
			queue[tail].l = x.l;
			queue[tail].r = x.r - 1;
			queue[tail].c = x.c;
			queue[tail++].depth = x.depth + 1;		
		}
		if(maze[x.l][x.r][x.c + 1] && !vis[x.l][x.r][x.c + 1]) {
			vis[x.l][x.r][x.c + 1] = true;
			queue[tail].l = x.l;
			queue[tail].r = x.r;
			queue[tail].c = x.c + 1;
			queue[tail++].depth = x.depth + 1;		
		}
		if(maze[x.l][x.r + 1][x.c] && !vis[x.l][x.r + 1][x.c]) {
			vis[x.l][x.r + 1][x.c] = true;
			queue[tail].l = x.l;
			queue[tail].r = x.r + 1;
			queue[tail].c = x.c;
			queue[tail++].depth = x.depth + 1;	
		}
		if(maze[x.l - 1][x.r][x.c] && !vis[x.l - 1][x.r][x.c]) {
			vis[x.l - 1][x.r][x.c] = true;
			queue[tail].l = x.l - 1; 
			queue[tail].r = x.r;
			queue[tail].c = x.c;
			queue[tail++].depth = x.depth + 1;	
		}
		if(maze[x.l + 1][x.r][x.c] && !vis[x.l + 1][x.r][x.c]) {
			vis[x.l + 1][x.r][x.c] = true;
			queue[tail].l = x.l + 1;
			queue[tail].r = x.r;
			queue[tail].c = x.c;
			queue[tail++].depth = x.depth + 1;	
		}
	}
	return false;
}
int main(int argc, char const *argv[])
{
	int L, R, C;
	while(cin >> L >> R >> C) {
		if(!L && !R && !C)
			break;
		memset(maze, false, sizeof(maze));
		for(int k = 1; k <= L; ++k)
			for(int i = 1; i <= R; ++i)
				for(int j = 1; j <= C; ++j) {
					char temp;
					cin >> temp;
					if(temp == '.')
						maze[k][i][j] = true;
					if(temp == 'S' ) {
						maze[k][i][j] = true;
						s.l = k;
						s.r = i;
						s.c = j;
					}
					if(temp == 'E') {
						maze[k][i][j] = true;
						e.l = k;
						e.r = i;
						e.c = j;
					}
				}
		if(bfs(s.l, s.r, s.c))
			cout << "Escaped in " << ans << " minute(s)." << endl;
		else
			cout << "Trapped!" << endl;
	}
	return 0;
}


非洲秃鹫优化算法(AVOA)详解:从原理到实战 非洲秃鹫优化算法(AVOA)是一种模拟秃鹫觅食行为的元启发式算法。该算法通过数学建模秃鹫的群体分工、饥饿感知和食物竞争等行为,将优化问题转化为智能觅食过程。AVOA包含四个关键阶段:(1)种群初始化与最佳秃鹫选择;(2)饥饿率控制探索与开发平衡;(3)探索阶段模拟远距离飞行搜索;(4)开发阶段模拟食物源附近的竞争行为。算法采用饥饿率函数实现自适应搜索,初期偏向全局探索,后期转向局部开发。实验表明AVOA在标准测试函数和工程设计问题上表现出优越性能,其创新性在于将秃鹫独特的生物学特性转化为高效的优化机制。 阅读详情

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% Dropout fraction opts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex4 neural net with sigmoid activation functionrand('state',0)nn = nnsetup([784 100 10]);nn.activation_function = 'sigm'; % Sigmoid activation functionnn.learningRate = 1; % Sigm require a lower learning rateopts.numepochs = 1; % Number of full sweeps through dataopts.batchsize = 100; % Take a mean gradient step over this many samplesnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex5 plotting functionalityrand('state',0)nn = nnsetup([784 20 10]);opts.numepochs = 5; % Number of full sweeps through datann.output = 'softmax'; % use softmax outputopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, train_x, train_y, opts);[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');%% ex6 neural net with sigmoid activation and plotting of validation and training error% split training data into training and validation datavx = train_x(1:10000,:);tx = train_x(10001:end,:);vy = train_y(1:10000,:);ty = train_y(10001:end,:);rand('state',0)nn = nnsetup([784 20 10]); nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

POJ - 2251 Dungeon Master (三维地图内的 bfs 找最短距离)

POJ - 2251 Dungeon Master (三维地图内的 bfs 找最短距离) 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...

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POJ 2251 Dungeon Master(BFS)

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POJ - 2251 - Dungeon Master(三维bfs)

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POJ2251 Dungeon Master BFS

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

colorfulshark 1081

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Tizzi的博客 873

POJ2251:Dungeon Master(BFS)

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

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超逸の学习技术博客 678

poj 2251 Dungeon Master 【BFS】

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shengweisong 627

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[kuangbin带你飞]专题1-23 专题一 简单搜索 POJ 2251 Dungeon Master 宽度优先遍历BFS+三维数组的使用

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mrcrack的博客 347
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