[장희직] - 케빈 베이컨의 6단계 법칙, 전구와 스위치, CCW, 함께 블록 쌓기 #172
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soopeach
approved these changes
Sep 11, 2023
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| val vectorP1P2 = Vector(p2X - p1X, p2Y - p1Y) | ||
| val vectorP1P3 = Vector(p3X - p1X, p3Y - p1Y) |
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| val queue = ArrayDeque<Pair<Int, Int>>() | ||
| queue.add(Pair(standard, 0)) | ||
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| while (queue.isNotEmpty()) { | ||
| val (man, count) = queue.removeFirst() | ||
| if (findRelations.add(man)) { | ||
| totalCount += count | ||
| relations[man].forEach { | ||
| if (it !in findRelations) queue.add(Pair(it, count + 1)) | ||
| } | ||
| } | ||
| } |
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아 저처럼 초기 친구들 안 구하고 이렇게 해도 되겠군요...ㅋㅋㅋㅋㅋㅋㅋㅋ
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| val dp = MutableList(size = h + 1) { 0 } | ||
| val tmpStored = MutableList(size = h + 1) { 0 } |
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1차원 dp 테이블 + 중복을 방지하기 위한 tmp저장소까지.. 다음에 비슷한 방식으로 시도해보겠습니다!
bngsh
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Sep 11, 2023
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| (vectorP1P2.x * vectorP1P3.y - vectorP1P3.x * vectorP1P2.y).run { | ||
| if (this < 0) print(-1) | ||
| if (this == 0) print(0) | ||
| if (this > 0) print(1) | ||
| } |
| answers[man] = bfs(man) | ||
| } | ||
| val minCount = answers.min() | ||
| println(answers) |
| } | ||
| val minCount = answers.min() | ||
| println(answers) | ||
| println(answers.indexOfFirst { it == minCount }) |
| dp[index] += i | ||
| dp[index] %= 10007 | ||
| } | ||
| tmpStored.fill(0) |
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헉...fill을 완전 잊고 있었네요 담엔 배열 초기화에 참고하겠습니다!
jeeminimini
approved these changes
Sep 11, 2023
| } | ||
| val minCount = answers.min() | ||
| println(answers) | ||
| println(answers.indexOfFirst { it == minCount }) |
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| state[0] = (state[0] + 1) % 2 | ||
| state[1] = (state[1] + 1) % 2 |
| dp[0] = 1 | ||
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| repeat(n) { | ||
| val heights = readln().split(' ').map { it.toInt() } |
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📌 from issue #170 📌
📋문제 목록📋
📝메모
CCW?