"use strict"; const {Fragment: _Fragment, jsx: _jsx, jsxs: _jsxs} = arguments[0]; const {useMDXComponents: _provideComponents} = arguments[0]; function _createMdxContent(props) { const _components = { a: "a", code: "code", figure: "figure", h2: "h2", img: "img", li: "li", p: "p", pre: "pre", span: "span", ul: "ul", ..._provideComponents(), ...props.components }; return _jsxs(_Fragment, { children: [_jsx(_components.p, { children: _jsx(_components.a, { href: "https://www.acmicpc.net/problem/11812", children: "문제 링크" }) }), "\n", _jsx(_components.h2, { children: "문제" }), "\n", _jsx(_components.p, { children: "각 노드가 자식을 최대 K개 가질 수 있는 트리를 K진 트리라고 한다. 총 N개의 노드로 이루어져 있는 K진 트리가 주어진다." }), "\n", _jsx(_components.p, { children: "트리는 \"적은 에너지\" 방법을 이용해서 만든다. \"적은 에너지\" 방법이란, 이전 깊이를 모두 채운 경우에만, 새로운 깊이를 만드는 것이고, 이 새로운 깊이의 노드는 가장 왼쪽부터 차례대로 추가 한다." }), "\n", _jsx(_components.p, { children: "아래 그림은 노드 9개로 이루어져 있는 3진 트리이다." }), "\n", _jsx(_components.p, { children: _jsx(_components.img, { src: "/generated/posts/algorithm/boj-11812/assets/img1.png", alt: "" }) }), "\n", _jsx(_components.p, { children: "노드의 개수 N과 K가 주어졌을 때, 두 노드 x와 y 사이의 거리를 구하는 프로그램을 작성하시오." }), "\n", _jsx(_components.h2, { children: "입력" }), "\n", _jsx(_components.p, { children: "첫째 줄에 N (1 <= N <= 10^15)과 K (1 <= K <= 1 000), 그리고 거리를 구해야 하는 노드 쌍의 개수 Q (1 <= Q <= 100 000)가 주어진다." }), "\n", _jsx(_components.p, { children: "다음 Q개 줄에는 거리를 구해야 하는 두 노드 x와 y가 주어진다. (1 <= x, y <= N, x ≠ y)" }), "\n", _jsx(_components.h2, { children: "출력" }), "\n", _jsx(_components.p, { children: "총 Q개의 줄을 출력한다. 각 줄에는 입력으로 주어진 두 노드 사이의 거리를 출력한다." }), "\n", _jsx(_components.h2, { children: "풀이" }), "\n", _jsxs(_components.p, { children: ["처음에는 트리를 실제로 만들어서 부모를 타고 올라가고 싶어지지만, ", _jsx(_components.code, { children: "N" }), "의 범위가 ", _jsx(_components.code, { children: "10^15" }), "라서 그 방법은 애초에 불가능하다. 그래서 현재 노드의 부모 번호를 직접 계산하는 방법이 필요하다."] }), "\n", _jsxs(_components.p, { children: ["이 트리는 각 레벨이 왼쪽부터 순서대로 채워지기 때문에, 노드 번호와 부모 번호 사이에 일정한 규칙이 있다. 루트를 1번이라고 하면, 어떤 노드 ", _jsx(_components.code, { children: "x" }), "의 부모는 ", _jsx(_components.code, { children: "(x - 2) / K + 1" }), "로 바로 구할 수 있다. 즉 트리를 만들지 않고도 부모 방향으로 올라가는 연산은 할 수 있다."] }), "\n", _jsxs(_components.p, { children: ["두 노드 사이의 거리는 결국 공통 조상에서 만날 때까지 더 큰 번호 쪽을 부모로 계속 올리면 구할 수 있다. 코드에서는 ", _jsx(_components.code, { children: "x != y" }), "인 동안 더 큰 쪽을 ", _jsx(_components.code, { children: "getParent()" }), "로 올리고, 한 번 올릴 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_jsx(_components.span, { style: { color: "#D8DEE9FF" }, children: " y" }), _jsx(_components.span, { style: { color: "#ECEFF4" }, children: ")" }), _jsx(_components.span, { style: { color: "#81A1C1" }, children: " <<" }), _jsx(_components.span, { style: { color: "#ECEFF4" }, children: " '" }), _jsx(_components.span, { style: { color: "#EBCB8B" }, children: "\\n" }), _jsx(_components.span, { style: { color: "#ECEFF4" }, children: "'" }), _jsx(_components.span, { style: { color: "#81A1C1" }, children: ";" })] }), "\n", _jsx(_components.span, { "data-line": "", children: _jsx(_components.span, { style: { color: "#ECEFF4" }, children: "\t}" }) }), "\n", _jsx(_components.span, { "data-line": "", children: _jsx(_components.span, { style: { color: "#D8DEE9FF" }, children: "\t" }) }), "\n", _jsxs(_components.span, { "data-line": "", children: [_jsx(_components.span, { style: { color: "#81A1C1" }, children: "\treturn" }), _jsx(_components.span, { style: { color: "#B48EAD" }, children: " 0" }), _jsx(_components.span, { style: { color: "#81A1C1" }, children: ";" })] }), "\n", _jsx(_components.span, { "data-line": "", children: _jsx(_components.span, { style: { color: "#ECEFF4" }, children: "}" }) })] }) }) }), "\n", _jsx(_components.h2, { children: "복잡도" }), "\n", _jsxs(_components.ul, { children: ["\n", _jsxs(_components.li, { children: ["시간 복잡도: 질의당 ", _jsx(_components.code, { children: "O(트리 높이)" })] }), "\n", _jsxs(_components.li, { children: ["공간 복잡도: ", _jsx(_components.code, { children: "O(1)" })] }), "\n"] }), "\n", _jsx(_components.h2, { children: "마무리" }), "\n", _jsx(_components.p, { children: "트리 문제라고 해서 항상 트리를 직접 만들 필요는 없다. 번호가 채워지는 규칙만 이용할 수 있다면, 부모를 수식으로 구해서 훨씬 가볍게 처리할 수 있다." }), "\n", _jsx(_components.p, { children: _jsx(_components.img, { src: "/generated/posts/algorithm/boj-11812/assets/1.png", alt: "" }) })] }); } function MDXContent(props = {}) { const {wrapper: MDXLayout} = { ..._provideComponents(), ...props.components }; return MDXLayout ? _jsx(MDXLayout, { ...props, children: _jsx(_createMdxContent, { ...props }) }) : _createMdxContent(props); } return { default: MDXContent };

BOJ 11812 - K진 트리