Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43132   Accepted: 17472 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..…

POJ 1458 Common Subsequence(LCS最长公共子序列)解题报告 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/F 题目: Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43388   Accepted: 17613 Description A subsequen…

解题思路:先注意到序列和串的区别,序列不需要连续,而串是需要连续的,先由样例abcfbc         abfcab画一个表格分析,用dp[i][j]储存当比较到s1[i],s2[j]时最长公共子序列的长度 a    b    f    c    a    b 0    0    0    0    0   0    0 a  0    1     1    1    1   1    1 b  0    1     2    2    2   2    2 c  0    1     2  …

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28494    Accepted Submission(s): 12735 Problem Description A subsequence of a given sequence is the given sequence with some e…

一.题目 Common Subsequence 二.分析 比较基础的求最长升序子序列. $DP[i][j]$表示的是字符串$S1[1...i]$与$S2[1...j]$的最长公共子序列长度. 状态转移:$$if s1[i] == s2[j]    DP[i][j] = DP[i-1][j-1] + 1$$  $$if s1[i] != s2[j]    DP[i][j] = max(DP[i-1][j], DP[i][j-1]$$ 相等时好理解,不相等的时候就是考虑两个字符串分别加上这个字符后,…

虽然以前可能接触过最长公共子序列,但是正规的写应该还是第一次吧. 直接贴代码就好了吧: #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; + ; char a[N],b[N]; int dp[N][N]; int main() { ,b+) == ) { ); ); memset(dp,,sizeof dp); ;i<=n;i++) { ;j<=m…

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][j]); (s[i] != t[i]) 代码: #include <cstdio> #include <cstring> #include <iostream> #in…

HDOJ 1159 Common Subsequence[DP] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44280 Accepted Submission(s): 20431 Problem Description A subsequence of a given sequence is the given sequence wit…

POJ1458 Common Subsequence(最长公共子序列LCS) http://poj.org/problem?id=1458 题意: 给你两个字符串, 要你求出两个字符串的最长公共子序列长度. 分析: 本题不用输出子序列,非常easy,直接处理就可以. 首先令dp[i][j]==x表示A串的前i个字符和B串的前j个字符的最长公共子序列长度为x. 初始化: dp全为0. 状态转移: IfA[i]==B[j] then dp[i][j]= dp[i-1][j-1]+1 else dp[…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1159 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28195#problem/A Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17621…