1. 问题描述 子串应该比较好理解,至于什么是子序列,这里给出一个例子:有两个母串 cnblogs belong 比如序列bo, bg, lg在母串cnblogs与belong中都出现过并且出现顺序与母串保持一致,我们将其称为公共子序列.最长公共子序列(Longest Common Subsequence, LCS),顾名思义,是指在所有的子序列中最长的那一个.子串是要求更严格的一种子序列,要求在母串中连续地出现.在上述例子的中,最长公共子序列为blog(cnblogs, belong),最长公…

Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a stri…

题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/F 题目: Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence…

''' merge two configure files, basic file is aFile insert the added content of bFile compare to aFile for example, 'bbb' is added content ----------------------------------------------------------- a file content | b file content | c merged file cont…

一.Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strict…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9595    Accepted Submission(s): 3923 Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18387    Accepted Submission(s): 7769 Problem Description A subsequence of a given sequence is the given sequence with some el…

Common Subsequence Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm&g…

2017-09-02 17:07:42 writer:pprp 通过这个题温习了一下刚学的LCS 代码如下: /* @theme:hdu1159 @writer:pprp @begin:17:01 @end:17:06 @declare:LCS的裸题,温习一下 @error:从1开始读入的话,用strlen也要从1开始测才可以 @date:2017/9/2 */ #include <bits/stdc++.h> using namespace std; ],s2[]; ][]; int mai…

最长公共子序列(LCS)是经典的DP问题,求序列a[1...n], b[1..m]的LCS. 状态是DP[i][j],表示a[1..i],b[1..j]的LCS. DP转移方程是 DP[i][j]= DP[i-1][j-1]+1, a[i] == b[j] max{ DP[i][j-1], DP[i-1][j] }, a[i] != b[i]  ----------------------------------------------------------------------------…