Baekjoon(C++)/BFS
[BOJ - 2573] 빙산(C++)
alswns8081
2025. 4. 29. 22:21
- cnt로 덩어리가 몇 개 있는지 확인(상하좌우 -> BFS)
- cnt == 1 -> 빙산이 하나만 존재
- cnt == 0 -> 모두 녹아버림 -> 0을 출력하면서 return 0(프로그램 끝내기)
- cnt > 1 -> 빙산이 분리된다
- visited 배열을 선언해서 0보다 큰 경우에만 true로 설정해서 들르도록 판별
코드
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int N, M;
vector<vector<int>> grid;
vector<vector<bool>> visited;
bool inArea(int x, int y) {
return x >= 0 && x < N && y >= 0 && y < M;
}
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> N >> M;
grid.resize(N, vector<int>(M));
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
cin >> grid[i][j];
}
int year = 0;
while (true) {
visited.assign(N, vector<bool>(M, false));
int cnt = 0;
queue<pair<int, int>> q;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (grid[i][j] != 0 && !visited[i][j]) {
q.push({i, j});
visited[i][j] = true;
cnt++;
}
while (!q.empty()) {
auto [cur_x, cur_y] = q.front();
q.pop();
for (int k = 0; k < 4; k++) {
int nx = cur_x + dx[k];
int ny = cur_y + dy[k];
if (!inArea(nx, ny)) continue;
if (visited[nx][ny]) continue;
if (grid[nx][ny] != 0) {
visited[nx][ny] = true;
q.push({nx, ny});
} else {
grid[cur_x][cur_y]--;
}
}
if (grid[cur_x][cur_y] < 0) grid[cur_x][cur_y] = 0;
}
}
}
if (cnt == 0) {
cout << 0;
return 0;
}
if (cnt > 1) break;
year++;
}
cout << year;
return 0;
}