"use strict"; const {Fragment: _Fragment, jsx: _jsx, jsxs: _jsxs} = arguments[0]; const {useMDXComponents: _provideComponents} = arguments[0]; function _createMdxContent(props) { const _components = { a: "a", annotation: "annotation", code: "code", figure: "figure", h2: "h2", img: "img", li: "li", math: "math", mi: "mi", mo: "mo", mrow: "mrow", ol: "ol", p: "p", pre: "pre", semantics: "semantics", span: "span", ul: "ul", ..._provideComponents(), ...props.components }; return _jsxs(_Fragment, { children: [_jsx(_components.p, { children: _jsx(_components.a, { href: "https://www.acmicpc.net/problem/1863", children: "문제 링크" }) }), "\n", _jsx(_components.h2, { children: "문제" }), "\n", _jsx(_components.p, { children: "도시에서 태양이 질 때에 보이는 건물들의 윤곽을 스카이라인이라고 한다. 스카이라인만을 보고서 도시에 세워진 건물이 몇 채인지 알아 낼 수 있을까? 건물은 모두 직사각형 모양으로 밋밋하게 생겼다고 가정한다." }), "\n", _jsx(_components.p, { children: "정확히 건물이 몇 개 있는지 알아내는 것은 대부분의 경우에 불가능하고, 건물이 최소한 몇 채 인지 알아내는 것은 가능해 보인다. 이를 알아내는 프로그램을 작성해 보자." }), "\n", _jsx(_components.h2, { children: "입력" }), "\n", _jsx(_components.p, { children: "첫째 줄에 n이 주어진다. (1 <= n <= 50,000) 다음 n개의 줄에는 왼쪽부터 스카이라인을 보아 갈 때 스카이라인의 고도가 바뀌는 지점의 좌표 x와 y가 주어진다. (1 <= x <= 1,000,000. 0 <= y <= 500,000) 첫 번째 지점의 x좌표는 항상 1이다." }), "\n", _jsx(_components.h2, { children: "출력" }), "\n", _jsx(_components.p, { children: "첫 줄에 최소 건물 개수를 출력한다." }), "\n", _jsx(_components.h2, { children: "풀이" }), "\n", _jsx(_components.p, { children: "문제 자체는 스택으로 해결하는 전형적인 유형이지만, 조건을 조금 꼼꼼히 봐야 한다." }), "\n", _jsxs(_components.p, { children: ["왼쪽에서 오른쪽으로 보면서 현재 유지 중인 건물 높이들을 스택에 넣어 관리한다. 새로 들어온 높이 ", _jsx(_components.code, { children: "y" }), "가 스택 top보다 작아지면 그보다 큰 높이들은 여기서 끝난 것이므로 pop 하면서 건물 개수를 증가시킨다."] }), "\n", _jsx(_components.p, { children: "주의할 점은 두 가지다." }), "\n", _jsxs(_components.ol, { children: ["\n", _jsx(_components.li, { children: "새 높이가 스택 top과 같다면 같은 높이의 건물이 이어지는 상황이므로 중복으로 세면 안 된다." }), "\n", _jsxs(_components.li, { children: ["새 높이가 ", _jsx(_components.code, { children: "0" }), "이라면 건물이 없는 구간을 뜻하므로 스택에 넣지 않아야 한다."] }), "\n"] }), "\n", _jsxs(_components.p, { children: ["즉, 높이가 내려갈 때만 pop 하며 개수를 세고, 같은 높이는 무시하고, ", _jsx(_components.code, { children: "0" 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encoding: "application/x-tex", children: "O(n)" })] }) }) }), _jsx(_components.span, { className: "katex-html", "aria-hidden": "true", children: _jsxs(_components.span, { className: "base", children: [_jsx(_components.span, { className: "strut", style: { height: "1em", verticalAlign: "-0.25em" } }), _jsx(_components.span, { className: "mord mathnormal", style: { marginRight: "0.0278em" }, children: "O" }), _jsx(_components.span, { className: "mopen", children: "(" }), _jsx(_components.span, { className: "mord mathnormal", children: "n" }), _jsx(_components.span, { className: "mclose", children: ")" })] }) })] })] }), "\n"] }), "\n", _jsx(_components.h2, { children: "마무리" }), "\n", _jsxs(_components.p, { children: ["스택 문제답게 구현은 짧지만, 같은 높이와 ", _jsx(_components.code, { children: "0" }), " 처리 조건을 빼먹으면 답이 쉽게 틀어진다. 핵심은 “언제 건물이 끝났다고 볼 것인가”를 스택의 pop 시점으로 정확히 잡는 데 있다."] }), "\n", _jsx(_components.p, { children: _jsx(_components.img, { src: "/generated/posts/algorithm/boj-1863/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 1863 - 스카이라인 쉬운거