백준 - 6593번 : 상범 빌딩 (C++)

풀이 방법 : BFS

3차원 배열이라는 것만 빼면 어느 BFS와 다를게 없는 문제
현재 위치에서 동서남북상하 6방향으로 이동할 수 있는지 체크하면서 목적지에 도달했을 경우 그 시간을 체크하면 된다.

#include <iostream>
#include <vector>
#include <queue>
#include <limits.h>

using namespace std;

struct Pos
{
	int x = 0;
	int y = 0;
	int z = 0;
	int Time = 0;
};

int DirX[6] = { 1,-1,0,0,0,0 };
int DirY[6] = { 0,0,1,-1,0,0 };
int DirZ[6] = { 0,0,0,0,1,-1 };

int main()
{
	while (true)
	{
		int L, R, C;

		cin >> L >> R >> C;

		if (L == 0 && R == 0 && C == 0)
			break;

		bool check[31][31][31] = {};
		char Building[31][31][31] = {};

		int StartX = 0, StartY = 0, StartZ = 0;
		int EndX = 0, EndY = 0, EndZ = 0;

		for (int i = 0; i < L; ++i)
		{
			for (int j = 0; j < R; ++j)
			{
				for (int k = 0; k < C; ++k)
				{
					cin >> Building[i][j][k];

					if (Building[i][j][k] == 'S')
					{
						StartX = k;
						StartY = j;
						StartZ = i;
					}

					else if (Building[i][j][k] == 'E')
					{
						EndX = k;
						EndY = j;
						EndZ = i;
					}
				}
			}
		}

		Pos Start;

		queue<Pos> pqPos;
		Start.x = StartX;
		Start.y = StartY;
		Start.z = StartZ;

		pqPos.push(Start);

		bool Enable = false;
		int MinTime = 0;

		while (!pqPos.empty())
		{
			Pos TopPos = pqPos.front();
			pqPos.pop();

			if (TopPos.x == EndX &&
				TopPos.y == EndY &&
				TopPos.z == EndZ)
			{
				Enable = true;
				MinTime = TopPos.Time;
				break;
			}


			if (check[TopPos.z][TopPos.y][TopPos.x])
				continue;

			check[TopPos.z][TopPos.y][TopPos.x] = true;

			for (int i = 0; i < 6; ++i)
			{
				Pos Next = TopPos;

				Next.x += DirX[i];
				Next.y += DirY[i];
				Next.z += DirZ[i];
				++Next.Time;

				if (Next.x < 0 || Next.x >= C ||
					Next.y < 0 || Next.y >= R ||
					Next.z < 0 || Next.z >= L)
					continue;

				if (Building[Next.z][Next.y][Next.x] == '#')
					continue;

				pqPos.push(Next);

			}
		}

		if (Enable)
			cout << "Escaped in " << MinTime << " minute(s).\n";

		else
			cout << "Trapped!\n";
	}

}