> For the complete documentation index, see [llms.txt](https://tobigs.gitbook.io/usaco-week2/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://tobigs.gitbook.io/usaco-week2/j.-bucket-brigade.md).

# J. Bucket Brigade

### 문제 설명 

B에서 L로 도달하는데 R만 피하면 됩니다.

여기서 R이나 L이 여러개면 BFS로 최단경로 탐색을 해야하지만

(문제 설명에 조건이 누락되어 있는데) B, L, R은 단 한개씩입니다.

그래서 B - R - L(또는 L - R - B)이 일직선으로 있는 경우만 예외처리해서 단순하게 구현하면 됩니다.

### 소스 코드 (C++)

```cpp
#include <iostream>
using namespace std;

int Bx, By, Rx, Ry, Lx, Ly, ans;

int main() {
	for (int i = 0; i < 10; i++) for (int j = 0; j < 10; j++) {
		char t;
		cin >> t;
		if (t == 'B') { Bx = i; By = j; }
		if (t == 'R') { Rx = i; Ry = j; }
		if (t == 'L') { Lx = i; Ly = j; }
	}

	ans = abs(Bx - Lx) + abs(By - Ly) - 1;

	if (Bx == Rx && Rx == Lx) {
		if (By < Ry && Ry < Ly)
			ans = 1 + abs(By - Ly);
		else if (Ry < By && Ly < Ry)
			ans = 1 + abs(By - Ly);
	}
		
	else if (By == Ry && Ry == Ly) {
		if (Bx < Rx && Rx < Lx)
			ans = 1 + abs(Bx - Lx);
		else if (Lx < Rx && Rx < Bx)
			ans = 1 + abs(Bx - Lx);
	}

	cout << ans << endl;
}
```


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