[소병희] - 내리막 길, 치즈, 가장 긴 증가하는 부분 수열 4, 삼각 달팽이 #270
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jeeminimini
approved these changes
May 26, 2024
| idx = j | ||
| } | ||
| } | ||
| if (idx != i) dp[i].addAll(dp[idx]) |
| var airContact = Array(n) { IntArray(m) } | ||
| val q = ArrayDeque<IntArray>() | ||
| var time = 0 | ||
| var cheese = 0 |
soopeach
approved these changes
May 26, 2024
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| if (idx != i) dp[i].addAll(dp[idx]) | ||
| dp[i].add(arr[i]) |
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오.. 요런식으로 하면 초기화 처리를 앞에 안해줘도 되는군요!
| return visited[r][c] | ||
| } | ||
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| println(dfs(0, 0)) |
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| while(side > 0) { | ||
| for(i in sl until sl + side) { | ||
| triangle[i].add(round, count++) | ||
| } | ||
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| repeat(side - 1) { i -> | ||
| triangle[sl + side - 1].add(round + i + 1, count++) | ||
| } | ||
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| for(i in sl + side-2 downTo sl + 1) { | ||
| triangle[i].add(triangle[i].size - round, count++) | ||
| } | ||
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| round++ | ||
| side -= 3 | ||
| sl += 2 | ||
| } |
| sl += 2 | ||
| } | ||
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| return triangle.flatMap { it.toList() }.toIntArray() |
jhg3410
approved these changes
May 26, 2024
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| for(i in 1 until n) { | ||
| var idx = i | ||
| for(j in i - 1 downTo 0) { | ||
| if (dp[j].last() < arr[i] && dp[j].size > dp[idx].size) { | ||
| idx = j | ||
| } | ||
| } |
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앞에서부터 탐색하는 게 가능할까 했는데 단순히 앞뒤 차이였군요,,
배열을 저장하는 것까지!
확실히 다른 풀이 보니깐 좋습니다!!👍
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| while(side > 0) { | ||
| for(i in sl until sl + side) { | ||
| triangle[i].add(round, count++) | ||
| } | ||
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| repeat(side - 1) { i -> | ||
| triangle[sl + side - 1].add(round + i + 1, count++) | ||
| } | ||
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| for(i in sl + side-2 downTo sl + 1) { | ||
| triangle[i].add(triangle[i].size - round, count++) | ||
| } | ||
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| round++ | ||
| side -= 3 | ||
| sl += 2 | ||
| } |
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오 중간에 넣어주는 것 시뮬레이션 생각하기 쉽지 않을 것 같은데,, 👍
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| fun checkAir() { | ||
| airContact = Array(n) { IntArray(m) } | ||
| airContact[0][0]++ | ||
| q.add(intArrayOf(0, 0)) | ||
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| while(q.isNotEmpty()) { | ||
| val (r, c) = q.removeFirst() | ||
| for(d in 0 until 4) { | ||
| val nr = r + dr[d] | ||
| val nc = c + dc[d] | ||
| if (nr !in 0 until n || nc !in 0 until m) continue | ||
| if (paper[nr][nc] == 0 && airContact[nr][nc] > 0) continue | ||
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| airContact[nr][nc]++ | ||
| if (paper[nr][nc] == 1) continue | ||
| q.add(intArrayOf(nr, nc)) | ||
| } | ||
| } | ||
| } |
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바깥의 공기들만 돌면서 치즈와 만나면 맞닿은 공기 개수까지 처리를 하는군요 👍🫡
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📌 from issue #266 📌
📋문제 목록📋