poj1458(LCS)c[
原创 2021-07-28 16:55:14
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POJ1458 Common Subsequence LCS模板题不说什么了直接上模板DescriptionA subsequence of a given sequence is the given sequen
原创 2023-06-12 14:15:44
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Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = another
转载 2013-07-29 20:02:00
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题目链接:http://poj.org/problem?id=1458 Common Subsequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 55099 Accepted: 22973 Description A
转载 2017-10-03 21:20:00
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Common SubsequenceTime Limit:1000MSMemory Limit:10000KTotal Submissions:43132Accepted:17472DescriptionA subsequence of a given sequence is the given s...
转载 2015-08-10 10:06:00
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题目传送门题意:输出两字符串的最长公共子序列长度分析:LCS(Longest Common Subsequence)裸题。状态转移方程:dp[i+1][j+1] = dp[i][j] + 1; (s[i] == t[i])dp[i+1][j+1] = max (dp[i][j+1], dp[i+1]...
转载 2015-03-23 19:29:00
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#include #include #include #include using namespace std;const int maxn = 5000;char 1))
原创 2022-08-17 15:35:07
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题目链接:http://poj.org/problem?id=1458 题目大意:给出两个
转载 2017-08-24 16:24:00
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基础DP。#include #include #include #include #include #include #include #include #include #include using namespace std;#define LL long long#defin...
转载 2016-07-06 10:06:00
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经典问题居然给忘了。。。#include <iostream>#include <string.h>using namespace std;//LCS问题int main(){ string s; string t; while(cin>>s>>t) { const char* a =...
原创 2022-01-30 16:54:51
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经典问题居然给忘了。。。#include <iostream>#include <string.h>using namespace std;//LCS问题int main(){ string s; string t; while(cin>>s>>t) { const char* a =...
原创 2021-07-13 16:33:41
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/* *问题描述:求最长公共子序列(POJ1458) */ #include #include using namesp
原创 2023-01-11 20:18:36
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题目链接:http://poj.org/problem?id=1458dp[i][j]dp[i] [j]表示串s1前i字符和串s2前j字符的最长公共子序列。#include<iostream>#include<string.h>#include<algorithm>using namespace std;int dp[1005][1005];char s1[1005];char s2[1
原创 2022-12-07 00:20:20
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Common SubsequenceTime Limit:1000MSMemory Limit:10000KTotal Submissions:39009Accepted:15713DescriptionA subsequence of a given sequence is the given s...
转载 2014-11-02 14:29:00
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题意:给定两个字符串,让你找出它们之间最长公共子序列(LCS)的长度。 析:很明显是个DP,就是LCS,一点都没变。设两个序列分别为,A1,A2,...和B1,B2..,d(i, j)表示两个字符串LCS长度。 当A[i] = B[j] 时,这个最长度就是上一个长度加1,即:d(i, j) = d(
介绍:LCS(最长公共子序列):给两个序列a,b,求出最大公共子序列的长度思路:dp[i][j] = (a[i] == b[j] ?  1 + dp[i - 1][j - 1]  :  max(dp[i - 1][j], dp[i][j - 1]));代码如下:/*POJ: 0ms,964KB*//*HDU: 31ms,2176KB*/#include
原创 2023-04-12 10:13:11
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// 题意:求两个字符串的最长公共子序列#include<iostream> //最长公共子序列#include<string>using namespace std;string str1,str2;int f[500][500];int lcs(int i,int j){ if(i==-1||j==-1) return 0; if(f[i][j]!=-1) return f[i][j]; else if(str1[i]==str2[j]) f[i][j]=lcs(i-1,j-1)+1; else ...
转载 2011-07-22 16:28:00
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Description A subsequence of a given sequence is the given sequence with some elements (possi…, zk &...
原创 2022-12-07 14:17:54
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题目大意:输入A,B字符串,A由a1,a2...an字符组成,B由b1,b2...bn字符组成。求最长公共子序列。解题思路:dp二维数组,dp[i][j]表示,a1,a2..ai与b1,b2...bj最长公共子序列。如果ai等于bj,dp[i][j] = dp[i-1][j-1] + 1;否则dp[i][j] = max(dp[i-1][j], dp[i][j-1]);ac代码:#include
原创 2021-12-01 15:10:55
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Common Subsequence POJ - 1458 求两序列的最长公共序列。 模板 L[i][j]=L[i-1][j-1]+1;更新 dio>#include<c...
原创 2022-08-22 21:18:42
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