[Silver I] Title: 징검다리 건너기 (small), Time: 16 ms, Memory: 2044 KB -Bae…

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KKanghh · Nov 5, 2023 · 83b6643 · 83b6643
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diff --git a/백준/Silver/22869. 징검다리 건너기 （small）/README.md b/백준/Silver/22869. 징검다리 건너기 （small）/README.md
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+# [Silver I] 징검다리 건너기 (small) - 22869 
+
+[문제 링크](https://www.acmicpc.net/problem/22869) 
+
+### 성능 요약
+
+메모리: 2044 KB, 시간: 16 ms
+
+### 분류
+
+다이나믹 프로그래밍
+
+### 제출 일자
+
+2023년 11월 5일 15:02:50
+
+### 문제 설명
+
+<p><mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"> <mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$N$</span></mjx-container>개의 돌이 일렬로 나열 되어 있다. <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$N$</span></mjx-container>개의 돌에는 왼쪽부터 차례대로 수 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-msub><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D434 TEX-I"></mjx-c></mjx-mi><mjx-script style="vertical-align: -0.15em;"><mjx-texatom size="s" texclass="ORD"><mjx-mn class="mjx-n"><mjx-c class="mjx-c31"></mjx-c></mjx-mn></mjx-texatom></mjx-script></mjx-msub><mjx-msub><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D434 TEX-I"></mjx-c></mjx-mi><mjx-script style="vertical-align: -0.15em;"><mjx-texatom size="s" texclass="ORD"><mjx-mn class="mjx-n"><mjx-c class="mjx-c32"></mjx-c></mjx-mn></mjx-texatom></mjx-script></mjx-msub><mjx-mo class="mjx-n"><mjx-c class="mjx-c2E"></mjx-c></mjx-mo><mjx-mo class="mjx-n" space="2"><mjx-c class="mjx-c2E"></mjx-c></mjx-mo><mjx-mo class="mjx-n" space="2"><mjx-c class="mjx-c2E"></mjx-c></mjx-mo><mjx-msub space="2"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D434 TEX-I"></mjx-c></mjx-mi><mjx-script style="vertical-align: -0.15em;"><mjx-texatom size="s" texclass="ORD"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D456 TEX-I"></mjx-c></mjx-mi></mjx-texatom></mjx-script></mjx-msub><mjx-mo class="mjx-n"><mjx-c class="mjx-c2E"></mjx-c></mjx-mo><mjx-mo class="mjx-n" space="2"><mjx-c class="mjx-c2E"></mjx-c></mjx-mo><mjx-mo class="mjx-n" space="2"><mjx-c class="mjx-c2E"></mjx-c></mjx-mo><mjx-msub space="2"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D434 TEX-I"></mjx-c></mjx-mi><mjx-script style="vertical-align: -0.15em;"><mjx-texatom size="s" texclass="ORD"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-texatom></mjx-script></mjx-msub></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>A</mi><mrow data-mjx-texclass="ORD"><mn>1</mn></mrow></msub><msub><mi>A</mi><mrow data-mjx-texclass="ORD"><mn>2</mn></mrow></msub><mo>.</mo><mo>.</mo><mo>.</mo><msub><mi>A</mi><mrow data-mjx-texclass="ORD"><mi>i</mi></mrow></msub><mo>.</mo><mo>.</mo><mo>.</mo><msub><mi>A</mi><mrow data-mjx-texclass="ORD"><mi>N</mi></mrow></msub></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$A_{1} A_{2} ... A_{i} ... A_{N}$</span></mjx-container>로 부여되어 있다. 가장 왼쪽에 있는 돌에서 출발하여 가장 오른쪽에 있는 돌로 건너가려고 한다.</p>
+
+<ol>
+	<li>항상 오른쪽으로만 이동할 수 있다.</li>
+	<li><mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"> <mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D456 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$i$</span></mjx-container>번째 돌에서 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D457 TEX-I"></mjx-c></mjx-mi><mjx-mo class="mjx-n"><mjx-c class="mjx-c28"></mjx-c></mjx-mo><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D456 TEX-I"></mjx-c></mjx-mi><mjx-mo class="mjx-n" space="4"><mjx-c class="mjx-c3C"></mjx-c></mjx-mo><mjx-mi class="mjx-i" space="4"><mjx-c class="mjx-c1D457 TEX-I"></mjx-c></mjx-mi><mjx-mo class="mjx-n"><mjx-c class="mjx-c29"></mjx-c></mjx-mo></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>j</mi><mo stretchy="false">(</mo><mi>i</mi><mo><</mo><mi>j</mi><mo stretchy="false">)</mo></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$j(i < j)$</span></mjx-container>번째 돌로 이동할 때 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mo class="mjx-n"><mjx-c class="mjx-c28"></mjx-c></mjx-mo><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D457 TEX-I"></mjx-c></mjx-mi><mjx-mo class="mjx-n" space="3"><mjx-c class="mjx-c2212"></mjx-c></mjx-mo><mjx-mi class="mjx-i" space="3"><mjx-c class="mjx-c1D456 TEX-I"></mjx-c></mjx-mi><mjx-mo class="mjx-n"><mjx-c class="mjx-c29"></mjx-c></mjx-mo></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mo stretchy="false">(</mo><mi>j</mi><mo>−</mo><mi>i</mi><mo stretchy="false">)</mo></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$(j - i)$</span></mjx-container> × (1 + |<mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-msub><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D434 TEX-I"></mjx-c></mjx-mi><mjx-script style="vertical-align: -0.15em;"><mjx-texatom size="s" texclass="ORD"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D456 TEX-I"></mjx-c></mjx-mi></mjx-texatom></mjx-script></mjx-msub><mjx-mo class="mjx-n" space="3"><mjx-c class="mjx-c2212"></mjx-c></mjx-mo><mjx-msub space="3"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D434 TEX-I"></mjx-c></mjx-mi><mjx-script style="vertical-align: -0.15em;"><mjx-texatom size="s" texclass="ORD"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D457 TEX-I"></mjx-c></mjx-mi></mjx-texatom></mjx-script></mjx-msub></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>A</mi><mrow data-mjx-texclass="ORD"><mi>i</mi></mrow></msub><mo>−</mo><msub><mi>A</mi><mrow data-mjx-texclass="ORD"><mi>j</mi></mrow></msub></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$A_{i} - A_{j}$</span></mjx-container>|) 만큼 힘을 쓴다.</li>
+	<li>돌을 한번 건너갈 때마다 쓸 수 있는 힘은 최대 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D43E TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$K$</span></mjx-container>이다.</li>
+</ol>
+
+<p>이때, 가장 왼쪽 돌에서 출발하여 가장 오른쪽에 있는 돌로 건너갈 수 있는지 구해보자.</p>
+
+### 입력 
+
+ <p>첫 번째 줄에 돌의 개수 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$N$</span></mjx-container>와 쓸 수 있는 최대 힘 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D43E TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>K</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$K$</span></mjx-container>가 공백으로 구분되어 주어진다.</p>
+
+<p>두 번째 줄에는 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D441 TEX-I"></mjx-c></mjx-mi></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$N$</span></mjx-container>개의 돌의 수 <mjx-container class="MathJax" jax="CHTML" style="font-size: 104.6%; position: relative;"><mjx-math class="MJX-TEX" aria-hidden="true"><mjx-msub><mjx-mi class="mjx-i"><mjx-c class="mjx-c1D434 TEX-I"></mjx-c></mjx-mi><mjx-script style="vertical-align: -0.15em;"><mjx-mi class="mjx-i" size="s"><mjx-c class="mjx-c1D456 TEX-I"></mjx-c></mjx-mi></mjx-script></mjx-msub></mjx-math><mjx-assistive-mml unselectable="on" display="inline"><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>A</mi><mi>i</mi></msub></math></mjx-assistive-mml><span aria-hidden="true" class="no-mathjax mjx-copytext">$A_i$</span></mjx-container>가 공백으로 구분되어 주어진다.</p>
+
+### 출력 
+
+ <p>가장 오른쪽에 있는 돌로 이동할 수 있다면 <code>YES</code>를 출력한다. 만약 이동하지 못하는 경우에는 <code>NO</code>를 출력한다.</p>
+
diff --git a/백준/Silver/22869. 징검다리 건너기 （small）/징검다리 건너기 （small）.cc b/백준/Silver/22869. 징검다리 건너기 （small）/징검다리 건너기 （small）.cc
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+#include <bits/stdc++.h>
+using namespace std;
+
+int A[5000];
+bool vis[5000];
+
+int main() {
+    ios::sync_with_stdio(0);
+    cin.tie(0);
+
+    int n, k;
+    cin >> n >> k;
+
+    for (int i = 0; i < n; i++) cin >> A[i];
+    vis[0] = true;
+
+    for (int i = 0; i < n; i++) {
+        if (!vis[i]) continue;
+        for (int j = i + 1; j < n; j++) {
+            if ((j - i) * (1 + abs(A[j] - A[i])) <= k) vis[j] = true;
+        }
+    }
+
+    if (vis[n - 1]) cout << "YES";
+    else cout << "NO";
+
+}