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🧷 문제 링크
https://www.acmicpc.net/problem/25189
🧭 풀이 시간
50분
👀 체감 난이도
✏️ 문제 설명
개구리가 격자 칸 안의 먹이를 먹는다. 그 먹이만큼 점프를 할 수 있는데 딱 1번 먹이를 무시하고 점프를 할 수 있다. 개구리가 집에 가기 위한 최소 점프 횟수를 구하라.
🔍 풀이 방법
visited[x][y][z]: x, y칸에 z(0: 무시점프 안함, 1: 무시점프 함)했을 때의 점프 횟수로 처리했다.
중요한 건 해당 행 혹은 열에서 무시점프를 했다면 다시 해당 행, 열에선 무시점프를 하면 안된다는 것이다. 이를 처리하지 않으면 시간초과가 발생했다.
⏳ 회고
오랜만에 생각해야 할 bfs문제라서 재밌었다.