Created
May 11, 2020 02:36
백준 문제풀이
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#include <bits/stdc++.h> | |
using namespace std; | |
// < z, x, y , time > | |
typedef tuple<int, int, int, int> tp; | |
int L, R, C; | |
int dx[] = { -1,1,0,0,0,0 }; | |
int dy[] = { 0,0,-1,1,0,0 }; | |
int dz[] = { 0,0,0,0,-1,1 }; | |
// arr[z][x][y] : 층(z), 행(x), 열(y) | |
string arr[33][33]; | |
bool visited[33][33][33]; | |
void CanOut(int a, int b, int c) { | |
queue<tp> Q; | |
visited[a][b][c] = true; | |
Q.push(tp(a, b, c, 0)); | |
while (!Q.empty()) { | |
tp now = Q.front(); | |
Q.pop(); | |
int z = get<0>(now); | |
int x = get<1>(now); | |
int y = get<2>(now); | |
int t = get<3>(now); | |
if (arr[z][x][y] == 'E') { | |
cout << "Escaped in " << t << " minute(s)." << '\n'; | |
return; | |
} | |
for (int i = 0; i < 6; ++i) { | |
int nz = z + dz[i]; | |
int nx = x + dx[i]; | |
int ny = y + dy[i]; | |
if (nz < 0 || nz >= L || nx < 0 || nx >= R || ny < 0 || ny >= C) continue; | |
if (visited[nz][nx][ny] || arr[nz][nx][ny] == '#') continue; | |
Q.push(tp(nz, nx, ny, t + 1)); | |
visited[nz][nx][ny] = true; | |
} | |
} | |
cout << "Trapped!" << '\n'; | |
} | |
int main() {ios_base::sync_with_stdio(false); cout.tie(NULL); cin.tie(NULL); | |
while (cin >> L >> R >> C) { | |
//초기화 | |
for (int i = 0; i < 30; ++i) | |
for (int j = 0; j < 30; ++j) | |
arr[i][j].clear(); | |
memset(visited, false, sizeof(visited)); | |
for (int z = 0; z < L; ++z) | |
for (int x = 0; x < R; ++x) | |
cin >> arr[z][x]; | |
for (int z = 0; z < L; ++z) | |
for (int x = 0; x < R; ++x) | |
for (int y = 0; y < C; ++y) | |
if (arr[z][x][y] == 'S') | |
CanOut(z, x, y); | |
} | |
return 0; | |
} |
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