CommonSubsequenceProblemDescriptionAsubsequenceofagivensequenceisthegivensequencewithsomeelements(possiblenone)leftout.GivenasequenceX=anothersequenceZ=isasubsequenceofXifthereexistsastrictlyincreasin

详情可看HDU1159。我们可以根据递推的方式得到最长的长度，如果想得到子序列也只需要记忆化一下每个点是从哪个点递推得到的即可。

longestcommonsub-sequences：最长-公共-子-序列子串：按原顺序依次出现，禁止跳过某元素子序列：在保持元素前后关系的前提下，可以跳过某些元素的序列例题：CommonSubsequence(HDU1159

这个题貌似是算法导论的原题不过本着能循环坚决不用递归的态度我认真地写了循环…关键其实就是一个转移方程如果两个字节一样，他就等于dp[a][b]=dp[a+1][b+1]+1;如果不一样就找两边最大的dp[a][b]=max(dp[a][b+1],dp[a+1][b]);另外在DP题里有点找到感觉了，如果想要用循环就先找到以前递归的感觉，然后自底向上加回去#include#include#inclu

dp1：最长上升子序列：hdu1950代码最长公共子序列hdu1159代码状压dp:TSP问题/货郎担问题hdu5418代码小练习：P1439【模板】最长公共子序列代码入门树形dphdu1520代码入门换根

坑的一笔，我脑残了，一直wa，原来的代码是从0开始遍历，这样if(a[i]==b[j])dp[i][j]=dp[i-1][[j-1]就会数组下标越界，所以经过修改才正确.附AC代码：#include#includechara[505];charb[505];intdp[505][505];intmax(intc,intd){returnc>d?c:d;}intmain(){intn,i,j,l1,

ProblemA：MaxSum（HDU1003）点击这里ProblemB：CommonSubsequence（HDU1159）点击这里ProblemC：SuperJumping!Jumping!

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):29329AcceptedSubmission(s):13174ProblemDescriptionAsubsequenceofagivensequenceisthegivenseque

HDU1159CommonSubsequence最长公共子序列（LCS）题目链接HDU1159解释这个是《算法竞赛从入门到进阶》的例题，就不是剽窃书上的讲解了。

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):57902AcceptedSubmission(s):26737ProblemDescriptionAsubsequenceofagivensequenceisthegivenseque

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):41128AcceptedSubmission(s):18995ProblemDescriptionAsubsequenceofagivensequenceisthegivenseque

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):40989AcceptedSubmission(s):18932ProblemDescriptionAsubsequenceofagivensequenceisthegivenseque

题目链接：hdu1159 点击打开链接题意：给你两个字符串，求lcs首先lcs的递推关系式是；d[i][j](以i为结尾的子序列1和以j为结尾的子序列2的最大公共子序列）        d[i-1][j

给出两字符串,要求输出两字符串的最长公共子序列;这里有必要解释一下子串和子序列的区别,子串要求连续,子序列不必.也就是说,对于"abckkkef"和"abcef"这两字符串而言,他们最长的公共子串是"abc",而最长公共子序列是"abcef"最长公共子序列,最长上升子序列,很显然都是动态规划的思想,不同的是,"公共"的话是要对两个序列进行处理mycode:#include #include #in

Asubsequenceofagivensequenceisthegivensequencewithsomeelements(possiblenone)leftout.GivenasequenceX=anothersequenceZ=isasubsequenceofXifthereexistsastrictlyincreasingsequenceofindicesofXsuchthatforall

题意：给出两个字符串，求最长公共子字符。如果s1[i]==s2[j],d[i][j]=d[i-1][j-1]+1,如果s1[i]!=s2[j],d[i][j]=max(d[i][j-1],d[i-1][j])。#include #include #include #include usingnamespacestd; intd[550][550]; intmain() { strings1,

 HDU1159题目大意：给定两个字符串，求他们的最长公共子序列的长度解题思路：设字符串a="a0,a1,a2,a3...am-1"(长度为m),b="b0,b1,b2,b3...bn-1"(长度为n)

importjava.util.Scanner; /** *HDU1159 *@authorashzheng *2016年3月17日17:48:29 */ publicclassLongestCommonSubsequence

hdu1159动态规划#include #include usingnamespacestd; inta[1001][1001]; intmain() { intlb,lc,i,j; charb[1001

坑的一笔，我脑残了，一直wa，原来的代码是从0开始遍历，这样if(a[i]==b[j])dp[i][j]=dp[i-1][[j-1]就会数组下标越界，所以经过修改才正确.附AC代码：#include #include chara[505]; charb[505]; intdp[505][505]; intmax(intc,intd) { returnc>d?c:d; } intmain() { i

http://acm.hdu.edu.cn/showproblem.php?pid=1159题意：给你两个字符串，求他们的最长公共子序列。思路：感觉比LIS略难，按照这个递推式理解吧：思路根源感觉和背包的放与不放相似。当相应字符相匹配时，一个两者都有的状态是由有你没我、有我没你、(两者都没然后+1)的状态转移而来的；当相应字符不匹配时，虽然(两者都没然后+1)的状态因为不匹配而无法加一，但是瘦死的

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)  MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):29956  AcceptedSubmission(s):13492ProblemDescriptionAsubsequenceofagivensequenceisthegivens

Common Subsequence Time Limit:1000MS      Memory Limit:32768KB      64bit IO Format:%I64d & %I64u Description A subsequence of a given se

这题实在是郁闷 检查了几个小时的错误 十几次runtime  居然是没定义成全局变量 m[][]存当前最大的公共子序列 m[i][j] = m[i-1][j-1] +1 (x[i] = y[i]) else m[i][j] = max(m[i-1][j],m[i][j-1]); View Code 1 #include<stdio.h> 2 #inc

最近一直在学习算法，基本上都是在学习动态规划以及字符串。当然，两者交集最经典之一则是LCS问题。 首先LCS的问题基本上就是在字符串a,b之间找到最长的公共子序列,比如 YAOLONGBLOG 和 YCLPBPG,其最长公共子序列则是YLBG 当然当字符串比较大时候，枚举则略显困难。 首先我们先考虑求一个基本问题，就是LCS的长度。 很容易可以理解递推式：  当a[i]==b[

题意：给两个字符串，求这两个字符串的最长公共子序列的长度 因为之前集训的时候做过，所以现在即使会做也并不是什么稀奇的事，依旧为了自己的浅薄感到羞愧啊``` 解法就是通过两个字符串的每个字符互相比较，根据比较情况相同与否确定递推关系： dp [ i + 1 ] [ j + 1 ] 表示匹配到 a 字符串的第 i 个字符和 b 字符串的第 j 个字符时的最大匹配数，由于读字符串的时候我是从下标

http://acm.hdu.edu.cn/showproblem.php?pid=1159 直接模板! View Code #include " iostream " using namespace std; char a[ 1001 ],

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,

题目描述： 　　就是求最长公共字串的长度，比如两个字符串abcd，abdef，那么它们的最长公共字串为abd， 　　长度为3。这个题目它不要求字串连续，其实连续也是一样的，两种dp而已。具体题目如下： 　　http://acm.hdu.edu.cn/showproblem.php?pid=1159 思路： 　　两个字符串s1,s2。我用dp[i][j]表示s1前i个字符与s2的前i个字符

Common Subsequence 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 sequ

LCS的无敌大水题#include #include #include #include #include #include usingnamespacestd; #definemaxn500 intdp[maxn][maxn]; charstr1[maxn],str2[maxn]; intmain() { inti,j; intm,n; while(~scanf("%s%s",str1+1,s

1 #include <iostream> 2 #include <cstring> 3 4 using namespace std; 5 6 int dp[1000][1000]; 7 char a[1000]; 8 char b[1000]; 9 10 int main() 11 { 12 while(cin>

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):29329    AcceptedSubmission(s):13174ProblemDescriptionAsubsequenceofagivensequenceisthegi

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):29176AcceptedSubmission(s):13084ProblemDescriptionAsubsequenceofagivensequenceisthegivenseque

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):29176AcceptedSubmission(s):13084ProblemDescriptionAsubsequenceofagivensequenceisthegivenseque

ProblemDescriptionAsubsequenceofagivensequenceisthegivensequencewithsomeelements(possiblenone)leftout.GivenasequenceX=anothersequenceZ=isasubsequenceofXifthereexistsastrictlyincreasingsequenceofindice

题目大意：给你两个字符串，让你找出他们的最长公共子序列。分析：裸LCS实现代码如下：#include #include #include #include usingnamespacestd; #definemaxn1005 intdp[maxn][maxn]; intmain() { stringstr1,str2; while(cin>>str1>>str2) { memset(dp,0,si

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):25465    AcceptedSubmission(s):11304ProblemDescriptionAsubsequenceofagivensequenceisthegi

CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):24919    AcceptedSubmission(s):11035ProblemDescriptionAsubsequenceofagivensequenceisthegi

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1159解题思路：基本的最长公共子序列，不懂的请看这个人的博客：http://www.cnblogs.com/xkfz007/archive/2012/10/17/2728728.htmlAC代码：#include #include #include usingnamespacestd; intdp[1

动态规划最长公共子序列a【】，b【】为两个已知序列，定义dp【i】【j】表示a【】的第i位和b【】的第j位之前的序列的最长公共子序列的长度。（1）a【i-1】==b【j-1】时，有dp【i】【j】=dp【i-1】【j-1】+1；（2）a【i-1】!=b【j-1】时，有dp【i】【j】=max（dp【i-1】【j】，dp【i】【j-1】）；#include #include #include #de

SDUT1299  数塔  HDU2084数塔  最长公共子序 HDU1159 CommonSubsequence  最大子段和 HDU1003MaxSum  编辑距离 SDUT1225编辑距离  杭电的

题目分析原题地址最简单的最长公共子序列（LCS）问题的模板题了。不解释。------------------------------------------------------------------------状态转移方程：dp[i][j]=dp[i-1][j-1]+1                             (a[i-1]==b[j-1])dp[i][j]=max(dp[i

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1159辅助空间变化示意图 abcfbca111111b122222f122333c123334a123334b123344由于f(i,j)只和f(i-1,j-1),f(i-1,j)和f(i,j-1)有关,而在计算f(i,j)时,只要选择一个合适的顺序,就可以保证这三项都已经计算出来了,这样就可以计算出f

CommonSubsequenceProblemDescriptionAsubsequenceofagivensequenceisthegivensequencewithsomeelements(possiblenone)leftout.GivenasequenceX=anothersequenceZ=isasubsequenceofXifthereexistsastrictlyincreasin

/*http://acm.hdu.edu.cn/showproblem.php?pid=1159CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)  MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):17607  AcceptedSubmission(s):7391ProblemDes

ProblemDescriptionAsubsequenceofagivensequenceisthegivensequencewithsomeelements(possiblenone)leftout.GivenasequenceX=anothersequenceZ=isasubsequenceofXifthereexistsastrictlyincreasingsequenceofindice

                                                   CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):17390    AcceptedSubmission(s):7290Pro

                                                   CommonSubsequenceTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):17390    AcceptedSubmission(s):7290Pro

#include #include #defineMAXN1002 intdp[MAXN][MAXN]; intmain() { charstr1[MAXN],str2[MAXN]; inti,j,len1,len2; while(scanf("%s%s",str1,str2)!=EOF) { memset(dp,0,sizeof(dp)); len1=strlen(str1); len2=s