(BOJ) 6593 상범 빌딩

문제: https://www.acmicpc.net/problem/6593

문제 풀이:

- 다익스트라 이용

- 동,서,남,북,상,하 이동

int dx[6] = { 0,0,1,0,-1,0 };

int dy[6] = { 0,0,0,1,0,-1 };

int dh[6] = { 1,-1,0,0,0,0 };

- tuple 이용해서 priority_queue에 저장

* tuple 설명

ㄴ 여러가지 변수를 묶을 수 있다 (pair랑 비슷)

ㄴ tuple<int,int,int> t = make_tuple(1,2,3);   // 값 삽입

ㄴ int a=get<0>(t); // t의 0번째 값 받아오기

소스 코드:

#include <iostream> #include <queue> #include <tuple> #include <vector> #include <string.h> #include <algorithm> #define inf 1000000000 using namespace std; int h, n, m,dist[30][30][30]; int dx[6] = { 0,0,1,0,-1,0 }; int dy[6] = { 0,0,0,1,0,-1 }; int dh[6] = { 1,-1,0,0,0,0 }; char arr[30][30][30]; bool visit[30][30][30]; int main() { ios::sync_with_stdio(false); cin.tie(NULL), cout.tie(NULL); while (1) { cin >> h >> n >> m; if (h == 0) return 0; memset(visit, false, sizeof(visit)); tuple<int, int, int> finish; priority_queue<tuple<int, int, int, int>> pq; for (int i = 0; i < h; i++) { for (int j = 0; j < n; j++) { for (int k = 0; k < m; k++) { cin >> arr[i][j][k]; dist[i][j][k] = inf; if (arr[i][j][k] == 'S') { dist[i][j][k] = 0; pq.push(make_tuple(0, i, j, k)); } else if (arr[i][j][k] == 'E') { finish = make_tuple(i, j, k); } } } } while (!pq.empty()) { int cur_h, cur_x, cur_y; do { auto t = pq.top(); cur_h = get<1>(t); cur_x = get<2>(t); cur_y = get<3>(t); pq.pop(); } while (visit[cur_h][cur_x][cur_y] && !pq.empty()); if (visit[cur_h][cur_x][cur_y]) break; visit[cur_h][cur_x][cur_y] = true; for (int i = 0; i < 6; i++) { int l = cur_h + dh[i], x = cur_x + dx[i], y = cur_y + dy[i]; if (l < 0 || l >= h || x < 0 || x >= n || y < 0 || y >= m || arr[l][x][y] == '#') continue; if (dist[l][x][y] > dist[cur_h][cur_x][cur_y] + 1) { dist[l][x][y] = dist[cur_h][cur_x][cur_y] + 1; pq.push({ -dist[l][x][y] ,l,x,y }); } } } if (dist[get<0>(finish)][get<1>(finish)][get<2>(finish)] == inf) cout << "Trapped!\n"; else cout << "Escaped in " << dist[get<0>(finish)][get<1>(finish)][get<2>(finish)] << " minute(s).\n"; } return 0; }