Common Subsequence

Time Limit: 2 Sec  Memory Limit: 64 MB
Submit: 951  Solved: 374

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 strictly increasing sequence <I1, ..., i2, ik>of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <A, b, f, c>is a subsequence of X = <A, b, f, c c,>with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

Input

The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.The length of the string is less than 1000.

Output

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc abfcab
programming contest
abcd mnp

Sample Output

4
2
0
 #include<stdio.h>
#include<string.h>
#define Max( a, b ) (a) > (b) ? (a) : (b) char s1[], s2[]; int dp[][]; int main()
{
int len1, len2;
while( scanf( "%s %s", s1+, s2+ ) != EOF )
{
memset( dp, , sizeof(dp) );
len1 = strlen( s1+ ), len2 = strlen( s2+ );
for( int i = ; i <= len1; ++i )
{
for( int j = ; j <= len2; ++j )
{
if( s1[i] == s2[j] )
{
dp[i][j] = dp[i-][j-] + ;
}
else
{
dp[i][j] = Max ( dp[i-][j], dp[i][j-] );
}
}
}
printf( "%d\n", dp[len1][len2] );
}
return ;
}

AC

Common Subsequence(dp)的更多相关文章

  1. UVA 10405 Longest Common Subsequence (dp + LCS)

    Problem C: Longest Common Subsequence Sequence 1: Sequence 2: Given two sequences of characters, pri ...

  2. POJ1458 Common Subsequence —— DP 最长公共子序列(LCS)

    题目链接:http://poj.org/problem?id=1458 Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Tot ...

  3. POJ - 1458 Common Subsequence DP最长公共子序列(LCS)

    Common Subsequence A subsequence of a given sequence is the given sequence with some elements (possi ...

  4. Longest Common Subsequence (DP)

    Given two strings, find the longest common subsequence (LCS). Your code should return the length of  ...

  5. HDU 1159 Common Subsequence --- DP入门之最长公共子序列

    题目链接 基础的最长公共子序列 #include <bits/stdc++.h> using namespace std; ; char c[maxn],d[maxn]; int dp[m ...

  6. CF 346B. Lucky Common Subsequence(DP+KMP)

    这题确实很棒..又是无想法..其实是AC自动机+DP的感觉,但是只有一个串,用kmp就行了. dp[i][j][k],k代表前缀为virus[k]的状态,len表示其他所有状态串,处理出Ac[len] ...

  7. POJ 1458 Common Subsequence DP

    http://poj.org/problem?id=1458 用dp[i][j]表示处理到第1个字符的第i个,第二个字符的第j个时的最长LCS. 1.如果str[i] == sub[j],那么LCS长 ...

  8. HDU 1159 Common Subsequence【dp+最长公共子序列】

    Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Other ...

  9. (线性dp,LCS) POJ 1458 Common Subsequence

    Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 65333   Accepted: 27 ...

随机推荐

  1. 20145215《Java程序设计》第6周学习总结

    20145215<Java程序设计>第六周学习总结 教材学习内容总结 输入/输出 InputStream与OutputStream 从应用程序角度来看,如果要将数据从来源取出,可以使用输入 ...

  2. 性能指标TP99之我解

    首先给出Google到的答案: The tp90 is a minimum time under which 90% of requests have been served. tp90 = top ...

  3. 从走出校门到Java实习生生活

    序 男,95年,这个学期就大四了,非计算机专业(数字媒体).目前在二线城市做Java实习生,待遇一般,应该算一个正常的实习生水平吧:租的一个约10平米的小单间,实习工资-衣食住行-杂七杂八的小消费差不 ...

  4. Python2和Python3 爬虫 转换

    由于Python3的不断完善,很多新入Python的小伙伴选择了Python3的阵营,很多人选择了爬虫这一热门话题,但是网络上大部分教程都是Python2 教程,Python3这一块做了些许的改动,对 ...

  5. Nodejs学习笔记(五)--- Express安装入门与模版引擎ejs

    目录 前言 Express简介和安装 运行第一个基于express框架的Web 模版引擎 ejs express项目结构 express项目分析 app.set(name,value) app.use ...

  6. 使用Web Deploy进行远程部署

    Web Deploy支持直接从本地Visual Studio的工程文件部署网站到远程服务器,部署的过程中可以对比哪些文件变化了需要拷贝,而不是一股脑的全部拷贝,效率和准确性会更好. 部署的过程主要要注 ...

  7. JavaEE EL的一些用法

    EL 可以在指示元素中设置EL是否使用 isELIgnored="true" true是不使用 也可以在web.xml中使用 <jsp-config> <jsp- ...

  8. iOS边练边学--NSURLSession、NSURLSessionTask的介绍与使用以及url中包含了中文的处理方法

    一.NSURLSession.NSURLSessionTask的使用步骤 首先创建NSURLSession对象 通过NSURLSession对象创建对应的任务 <1>NSURLSessio ...

  9. HashMap 和 HashTable区别

    HashMap 非线程安全的 HashTable线程安全的 package Collections.Map; import java.util.HashMap; public class HashMa ...

  10. 问问题_为什么关闭浏览器后Session会失效

    首先需要理解一下几点: 1.Http是无状态的,即对于每一次请求都是一个全新的请求,服务器不保存上一次请求的信息 2.Session是保存在服务端的,为什么后续请求会读取到session?因为请求会包 ...