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[2023-09-11] wooyeol #205#219

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[2023-09-11] wooyeol #205 #219

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46 changes: 46 additions & 0 deletions BOJ/LCS/wooyeol.py
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"""
LCS
https://www.acmicpc.net/problem/9251

풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

풀이시간
13:22 ~ 14:22(1시간 고민 후 풀이 참고)

문제 조건
문자열의 최대 길이는 1000

시간 복잡도 :
O(N * N) = O(N^2)

접근법
무슨 알고리즘으로 풀이 할 수 있을까? -> DP

점화식:
1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
for j in range(1, s2_length + 1):
# 점화식을 코드로 옮긴 내용
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])
, 'i'); if (__m === '*' || __re.test(location.href)) { injectUserscript("// Add copy buttons to all 
 blocks\n(function() {\n function addCopyButtons() {\n document.querySelectorAll('pre code').forEach(function(codeBlock) {\n if (codeBlock.parentElement.hasAttribute('data-copy-added')) return;\n codeBlock.parentElement.setAttribute('data-copy-added', 'true');\n \n var btn = document.createElement('button');\n btn.textContent = 'Copy';\n btn.style.cssText = 'position:absolute;top:4px;right:4px;padding:2px 8px;font-size:11px;background:#4ecdc4;border:none;border-radius:4px;color:#1a1a2e;cursor:pointer;opacity:0.7;transition:opacity 0.2s;';\n btn.onmouseover = function() { this.style.opacity = '1'; };\n btn.onmouseout = function() { this.style.opacity = '0.7'; };\n btn.onclick = function() {\n navigator.clipboard.writeText(codeBlock.textContent).then(function() {\n btn.textContent = 'Copied!';\n setTimeout(function() { btn.textContent = 'Copy'; }, 1500);\n });\n };\n codeBlock.parentElement.style.position = 'relative';\n codeBlock.parentElement.appendChild(btn);\n });\n }\n \n addCopyButtons();\n \n // Re-run on dynamic content\n var observer = new MutationObserver(addCopyButtons);\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Add Copy Buttons to Code Blocks");
}
} catch(__e) { console.warn('[Userscript:Add Copy Buttons to Code Blocks]', __e); }
})();
(function(){
try {
var __m = "github.com";
var __re = new RegExp('^' + "github\\.com" + '

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[2023-09-11] wooyeol #205#219
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merged 1 commit into

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Sep 11, 2023
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[2023-09-11] wooyeol #205
#219
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46 changes: 46 additions & 0 deletions
BOJ/LCS/wooyeol.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,46 @@
"""
 LCS
 https://www.acmicpc.net/problem/9251

 풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

 풀이시간
 13:22 ~ 14:22(1시간 고민 후 풀이 참고)

 문제 조건
 문자열의 최대 길이는 1000

 시간 복잡도 :
 O(N * N) = O(N^2)

 접근법
 무슨 알고리즘으로 풀이 할 수 있을까? -> DP

 점화식:
 1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
 2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
 3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
 4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
 for j in range(1, s2_length + 1):
 # 점화식을 코드로 옮긴 내용
 if s1[i - 1] == s2[j - 1]:
 dp[i][j] = dp[i - 1][j - 1] + 1
 else:
 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Force GitHub README to respect dark mode\n(function() {\n var style = document.createElement('style');\n style.textContent = '\n .markdown-body {\n color-scheme: dark light;\n }\n .markdown-body pre { background: #161b22 !important; }\n .markdown-body code { background: rgba(110, 118, 129, 0.4) !important; }\n .markdown-body table th, .markdown-body table td { border-color: #30363d !important; }\n .markdown-body img { background: #0d1117; }\n .markdown-body blockquote { border-left-color: #8b949e; }\n .markdown-body hr { border-color: #30363d; }\n ';\n document.head.appendChild(style);\n})();", "GitHub Dark Mode README Fix");
}
} catch(__e) { console.warn('[Userscript:GitHub Dark Mode README Fix]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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[2023-09-11] wooyeol #205#219
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Woo-Yeol
merged 1 commit into

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Sep 11, 2023
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[2023-09-11] wooyeol #205
#219
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46 changes: 46 additions & 0 deletions
BOJ/LCS/wooyeol.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,46 @@
"""
 LCS
 https://www.acmicpc.net/problem/9251

 풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

 풀이시간
 13:22 ~ 14:22(1시간 고민 후 풀이 참고)

 문제 조건
 문자열의 최대 길이는 1000

 시간 복잡도 :
 O(N * N) = O(N^2)

 접근법
 무슨 알고리즘으로 풀이 할 수 있을까? -> DP

 점화식:
 1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
 2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
 3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
 4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
 for j in range(1, s2_length + 1):
 # 점화식을 코드로 옮긴 내용
 if s1[i - 1] == s2[j - 1]:
 dp[i][j] = dp[i - 1][j - 1] + 1
 else:
 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Highlight search terms from Google/DuckDuckGo/Bing referrer\n(function() {\n var ref = document.referrer;\n var terms = [];\n \n if (ref.includes('google.com') || ref.includes('duckduckgo.com') || ref.includes('bing.com')) {\n var url = new URL(ref);\n var q = url.searchParams.get('q') || url.searchParams.get('p');\n if (q) {\n terms = q.split(/\\s+/).filter(function(t) { return t.length > 2; });\n }\n }\n \n if (terms.length === 0) return;\n \n var style = document.createElement('style');\n style.textContent = '.userscript-highlight { background: #fbbf24; color: #1a1a2e; padding: 1px 3px; border-radius: 2px; }';\n document.head.appendChild(style);\n \n function highlight(node) {\n if (node.nodeType === 3) { // text node\n var text = node.textContent;\n var found = false;\n terms.forEach(function(term) {\n var regex = new RegExp('(' + term.replace(/[.*+?^${}()|[\\]\\\\]/g, '\\\\') + ')', 'gi');\n if (regex.test(text)) {\n found = true;\n var frag = document.createDocumentFragment();\n var parts = text.split(regex);\n parts.forEach(function(part, i) {\n if (i % 2 === 0) {\n frag.appendChild(document.createTextNode(part));\n } else {\n var span = document.createElement('span');\n span.className = 'userscript-highlight';\n span.textContent = part;\n frag.appendChild(span);\n }\n });\n node.parentNode.replaceChild(frag, node);\n }\n });\n } else if (node.nodeType === 1 && node.childNodes) { // element\n var skipTags = ['SCRIPT', 'STYLE', 'NOSCRIPT', 'TEXTAREA', 'INPUT', 'SELECT'];\n if (!skipTags.includes(node.tagName)) {\n Array.from(node.childNodes).forEach(highlight);\n }\n }\n }\n \n highlight(document.body);\n \n // Re-highlight on dynamic content\n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1 || node.nodeType === 3) highlight(node);\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Highlight Search Terms");
}
} catch(__e) { console.warn('[Userscript:Highlight Search Terms]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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[2023-09-11] wooyeol #205#219
 Merged

Woo-Yeol
merged 1 commit into

from

Sep 11, 2023
 Merged


[2023-09-11] wooyeol #205
#219
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46 changes: 46 additions & 0 deletions
BOJ/LCS/wooyeol.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,46 @@
"""
 LCS
 https://www.acmicpc.net/problem/9251

 풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

 풀이시간
 13:22 ~ 14:22(1시간 고민 후 풀이 참고)

 문제 조건
 문자열의 최대 길이는 1000

 시간 복잡도 :
 O(N * N) = O(N^2)

 접근법
 무슨 알고리즘으로 풀이 할 수 있을까? -> DP

 점화식:
 1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
 2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
 3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
 4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
 for j in range(1, s2_length + 1):
 # 점화식을 코드로 옮긴 내용
 if s1[i - 1] == s2[j - 1]:
 dp[i][j] = dp[i - 1][j - 1] + 1
 else:
 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Strip utm_, fbclid, gclid, etc. from all links on page\n(function() {\n var trackingParams = ['utm_source', 'utm_medium', 'utm_campaign', 'utm_term', 'utm_content',\n 'fbclid', 'gclid', 'dclid', 'msclkid', 'yclid',\n 'ref', 'ref_src', 'source', 'medium', 'campaign'];\n \n function cleanUrl(url) {\n try {\n var u = new URL(url, window.location.origin);\n var changed = false;\n trackingParams.forEach(function(p) {\n if (u.searchParams.has(p)) {\n u.searchParams.delete(p);\n changed = true;\n }\n });\n return changed ? u.toString() : url;\n } catch (e) {\n return url;\n }\n }\n \n function cleanLinks() {\n document.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n \n cleanLinks();\n \n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1) {\n if (node.tagName === 'A') cleanLinks();\n node.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Remove Tracking Parameters from Links");
}
} catch(__e) { console.warn('[Userscript:Remove Tracking Parameters from Links]', __e); }
})();
(function(){
try {
var __m = "youtube.com";
var __re = new RegExp('^' + "youtube\\.com" + '

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[2023-09-11] wooyeol #205#219
 Merged

Woo-Yeol
merged 1 commit into

from

Sep 11, 2023
 Merged


[2023-09-11] wooyeol #205
#219
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46 changes: 46 additions & 0 deletions
BOJ/LCS/wooyeol.py

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Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,46 @@
"""
 LCS
 https://www.acmicpc.net/problem/9251

 풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

 풀이시간
 13:22 ~ 14:22(1시간 고민 후 풀이 참고)

 문제 조건
 문자열의 최대 길이는 1000

 시간 복잡도 :
 O(N * N) = O(N^2)

 접근법
 무슨 알고리즘으로 풀이 할 수 있을까? -> DP

 점화식:
 1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
 2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
 3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
 4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
 for j in range(1, s2_length + 1):
 # 점화식을 코드로 옮긴 내용
 if s1[i - 1] == s2[j - 1]:
 dp[i][j] = dp[i - 1][j - 1] + 1
 else:
 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Auto-enable theater mode on YouTube\n(function() {\n function tryTheater() {\n var btn = document.querySelector('button[aria-label=\"Theater mode\"], ytd-player #player button[title=\"Theater mode\"]');\n if (btn && !btn.classList.contains('activated')) {\n btn.click();\n }\n }\n \n // Try immediately\n tryTheater();\n \n // Try after navigation (SPA)\n var lastUrl = location.href;\n setInterval(function() {\n if (location.href !== lastUrl) {\n lastUrl = location.href;\n setTimeout(tryTheater, 500);\n }\n }, 1000);\n \n // Also try on player load\n var observer = new MutationObserver(tryTheater);\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "YouTube Theater Mode Default");
}
} catch(__e) { console.warn('[Userscript:YouTube Theater Mode Default]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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[2023-09-11] wooyeol #205#219
 Merged

Woo-Yeol
merged 1 commit into

from

Sep 11, 2023
 Merged


[2023-09-11] wooyeol #205
#219
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46 changes: 46 additions & 0 deletions
BOJ/LCS/wooyeol.py

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Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,46 @@
"""
 LCS
 https://www.acmicpc.net/problem/9251

 풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

 풀이시간
 13:22 ~ 14:22(1시간 고민 후 풀이 참고)

 문제 조건
 문자열의 최대 길이는 1000

 시간 복잡도 :
 O(N * N) = O(N^2)

 접근법
 무슨 알고리즘으로 풀이 할 수 있을까? -> DP

 점화식:
 1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
 2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
 3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
 4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
 for j in range(1, s2_length + 1):
 # 점화식을 코드로 옮긴 내용
 if s1[i - 1] == s2[j - 1]:
 dp[i][j] = dp[i - 1][j - 1] + 1
 else:
 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Remove or un-stick sticky/fixed headers that block content\n(function() {\n function unstick() {\n document.querySelectorAll('header, nav, [role=\"banner\"], .header, .navbar, .sticky, .fixed-top, [style*=\"position: fixed\"], [style*=\"position:sticky\"]').forEach(function(el) {\n if (el.style.position === 'fixed' || el.style.position === 'sticky' || \n getComputedStyle(el).position === 'fixed' || getComputedStyle(el).position === 'sticky') {\n el.style.position = 'static';\n el.style.top = 'auto';\n el.style.zIndex = 'auto';\n }\n });\n }\n \n unstick();\n \n var observer = new MutationObserver(unstick);\n observer.observe(document.body, { childList: true, subtree: true, attributes: true, attributeFilter: ['style', 'class'] });\n})();", "Kill Sticky Headers");
}
} catch(__e) { console.warn('[Userscript:Kill Sticky Headers]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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[2023-09-11] wooyeol #205#219
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Sep 11, 2023
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[2023-09-11] wooyeol #205
#219
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BOJ/LCS/wooyeol.py

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"""
 LCS
 https://www.acmicpc.net/problem/9251

 풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

 풀이시간
 13:22 ~ 14:22(1시간 고민 후 풀이 참고)

 문제 조건
 문자열의 최대 길이는 1000

 시간 복잡도 :
 O(N * N) = O(N^2)

 접근법
 무슨 알고리즘으로 풀이 할 수 있을까? -> DP

 점화식:
 1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
 2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
 3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
 4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
 for j in range(1, s2_length + 1):
 # 점화식을 코드로 옮긴 내용
 if s1[i - 1] == s2[j - 1]:
 dp[i][j] = dp[i - 1][j - 1] + 1
 else:
 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])

, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Universal Dark Mode - works on any site\n(function() {\n var enabled = true;\n \n function applyDarkMode() {\n if (!enabled) return;\n \n // Create style element if it doesn't exist\n var style = document.getElementById('universal-dark-mode-style');\n if (!style) {\n style = document.createElement('style');\n style.id = 'universal-dark-mode-style';\n document.head.appendChild(style);\n }\n \n // Dark mode CSS - inverts colors but preserves images/video\n style.textContent = '\n /* Invert everything except media */\n html {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #1a1a2e !important;\n }\n \n /* Restore images, videos, iframes, canvas */\n img, video, iframe, canvas, svg, picture, [style*=\"background-image\"] {\n filter: invert(1) hue-rotate(180deg) !important;\n }\n \n /* Preserve specific elements that should not be inverted */\n .no-dark-mode, .no-dark-mode *,\n [data-theme=\"light\"], [data-theme=\"light\"],\n .ace_editor, .ace_editor *,\n .CodeMirror, .CodeMirror *,\n .monaco-editor, .monaco-editor *,\n .markdown-body pre, .markdown-body pre *,\n .highlight, .highlight *,\n pre code, pre code * {\n filter: none !important;\n }\n \n /* Fix common UI elements */\n .modal, .popup, .dropdown-menu, .tooltip, .popover {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #2d2d44 !important;\n border-color: #444 !important;\n }\n \n /* Scrollbars */\n ::-webkit-scrollbar { background: #1a1a2e !important; }\n ::-webkit-scrollbar-thumb { background: #444 !important; }\n ::-webkit-scrollbar-thumb:hover { background: #555 !important; }\n \n /* Selection */\n ::selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ::-moz-selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ';\n }\n \n function removeDarkMode() {\n var style = document.getElementById('universal-dark-mode-style');\n if (style) style.remove();\n }\n \n // Toggle with Alt+Shift+D\n document.addEventListener('keydown', function(e) {\n if (e.altKey && e.shiftKey && e.key === 'D') {\n e.preventDefault();\n enabled = !enabled;\n if (enabled) {\n applyDarkMode();\n console.log('[Universal Dark Mode] Enabled');\n } else {\n removeDarkMode();\n console.log('[Universal Dark Mode] Disabled');\n }\n }\n });\n \n // Apply on load\n applyDarkMode();\n \n // Re-apply on dynamic content\n var observer = new MutationObserver(function(mutations) {\n if (enabled && !document.getElementById('universal-dark-mode-style')) {\n applyDarkMode();\n }\n });\n observer.observe(document.head, { childList: true });\n \n console.log('[Universal Dark Mode] Loaded - Press Alt+Shift+D to toggle');\n})();", "Universal Dark Mode");
}
} catch(__e) { console.warn('[Userscript:Universal Dark Mode]', __e); }
})();
})();

Skip to content

This repository was archived by the owner on Mar 18, 2024. It is now read-only.

[2023-09-11] wooyeol #205#219
 Merged

Woo-Yeol
merged 1 commit into

from

Sep 11, 2023
 Merged


[2023-09-11] wooyeol #205
#219
 Changes from all commits
Commits

Show all changes


1 commit


Select commit


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Filter by extension
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46 changes: 46 additions & 0 deletions
BOJ/LCS/wooyeol.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,46 @@
"""
 LCS
 https://www.acmicpc.net/problem/9251

 풀이 참조 : https://velog.io/@emplam27/알고리즘-그림으로-알아보는-LCS-알고리즘-Longest-Common-Substring와-Longest-Common-Subsequence

 풀이시간
 13:22 ~ 14:22(1시간 고민 후 풀이 참고)

 문제 조건
 문자열의 최대 길이는 1000

 시간 복잡도 :
 O(N * N) = O(N^2)

 접근법
 무슨 알고리즘으로 풀이 할 수 있을까? -> DP

 점화식:
 1. 문자열A, 문자열B의 한글자씩 비교해봅니다.
 2. 두 문자가 다르다면 LCS[i - 1][j]와 LCS[i][j - 1] 중에 큰값을 표시합니다.
 3. 두 문자가 같다면 LCS[i - 1][j - 1] 값을 찾아 +1 합니다.
 4. 위 과정을 반복합니다.

"""
import sys

input = sys.stdin.readline

s1 = input().rstrip()
s2 = input().rstrip()

s1_length = len(s1)
s2_length = len(s2)

dp = [[0] * (s2_length + 1) for _ in range(s1_length + 1)]

for i in range(1, s1_length + 1):
 for j in range(1, s2_length + 1):
 # 점화식을 코드로 옮긴 내용
 if s1[i - 1] == s2[j - 1]:
 dp[i][j] = dp[i - 1][j - 1] + 1
 else:
 dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

print(dp[-1][-1])