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🧷 문제 링크
https://www.acmicpc.net/problem/13023
🧭 풀이 시간
25분
👀 체감 난이도
✏️ 문제 설명
친구관계가 주어질 때 A->B->C->D->E 처럼 5명이 연속으로 친구가 되는 경우에 1 앞선 경우가 없으면 0을 출력하는 문제입니다.
🔍 풀이 방법
dfs를 이용하여 풀었습니다.
5명이 연속해서 친구가 되면 플래그를 true로 하여 이를 판단하여 출력하였습니다.
⏳ 회고
n이 2000이었기 때문에 최대 2억번 연산이고 시간제한이 2초라서 인접행렬로 풀었는데 시간초과가 발생해서 인접리스트로 바꿨습니다.
문제 조건을 잘 파악하여 사용해야겠습니다.