# 16234-인구이동

> [www.acmicpc.net/problem/16234](https://www.acmicpc.net/problem/16234)

1. 인구이동을 실시한다.
2. 인구이동을 실시하는 와중에 인구이동이 이뤄진다면 swich = ' on ' 해준다.
3. 인구이동이 끝난 후 swich 가 ' on ' 이라면 cnt += 1 해주고 또 실시한다.

```python
def dfs(y,x):
    global n,l,r,cnt,swich
    # 상하좌우에 대해서 탐색
    for i in range(4):
        ny = y + dy[i]
        nx = x + dx[i]
        # 인덱싱 범위 안이고, 인구차이 수가 문제조건 해당이고, 방문하지 않았다면
        if 0 <= ny < n and 0 <= nx < n and l <= abs(arr[y][x] - arr[ny][nx]) <= r and not visited[ny][nx]:
            swich = 'on'
            locs.append((ny,nx))
            cnt += arr[ny][nx]
            visited[ny][nx] = True
            dfs(ny,nx)
    return

n,l,r = map(int,input().split())
arr = [list(map(int,input().split())) for _ in range(n)]

visited = [[False for _ in range(n)] for __ in range(n)]
dy = [-1,1,0,0]
dx = [0,0,-1,1]

swich = 'on'
rst = 0

while swich == 'on':
    swich = 'off'
    visited = [[False for _ in range(n)] for __ in range(n)]
    for y in range(n):
        for x in range(n):
            locs = []
            if not visited[y][x]:
                visited[y][x] = True
                cnt = arr[y][x]
                locs.append((y,x))
                dfs(y,x)

                for ny,nx in locs:
                    arr[ny][nx] = cnt//len(locs)
    if swich == 'on':
        rst += 1

print(rst)
```


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