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. VmWare Workstation 10 安装 Ubuntu 14.04 问题解决

    Ubuntu安装过程很顺利,安装完成后还是有小问题存在   问题1:无法联网,PING可以通,网址无法解析 原因:默认DNS设置不正确 解决:设置DNS地址为8.8.8.8,问题解决   问题2:vm ...

  2. Django实际站点项目开发经验谈

    开发了两个月的Django站点正式上线了,看着网站从无到有,从前端到后台,从本地开发到环境部署,一点一滴的堆砌成型,着实带给我不小的乐趣. Django站点介绍: 开发环境:阿里云服务器centos6 ...

  3. JavaScript实现Ajax小结

    置顶文章:<纯CSS打造银色MacBook Air(完整版)> 上一篇:<TCP的三次握手和四次挥手> 作者主页:myvin 博主QQ:851399101(点击QQ和博主发起临 ...

  4. 第十六章:脚本化HTTP

    写在本章内容前: 第十五章:事件处理 涉及到到较多的文字篇幅,介于个人精力问题,暂不更新.主要包含的内容有事件类型.注册事件处理程序.事件处理程序的调用.文档加载事件.鼠标事件.鼠标滚轮事件.拖放事件 ...

  5. AngularJS - 服务简介

    服务是AngularJS中非常重要的一个概念,虽然我们有了控制器,但考虑到其生命实在脆弱,我们需要用到服务. 起初用service时,我便把service和factory()理所当然地关联起来了. 确 ...

  6. 【Moqui业务逻辑翻译系列】Shipment Receiver Receives Shipment with Packing Slip but no PO

    Shipment Receiver receives shipment. It has invoice tucked into it. Receiver records vendor name, ve ...

  7. 【 CodeForces 604A】B - 特别水的题2-Uncowed Forces

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/B Description Kevin Sun has jus ...

  8. 【HDU 5363】Key Set

    题 Description soda has a set $S$ with $n$ integers $\{1, 2, \dots, n\}$. A set is called key set if ...

  9. RBAC(基于角色的访问控制权限)表结构

    Rbac 支持两种类,PhpManager(基于文件的) 和 DbManager(基于数据库的) 权限:就是指用户是否可以执行哪些操作 角色:就是上面说的一组操作的集合,角色还可以继承 在Yii2.0 ...

  10. 【poj1182】 食物链

    http://poj.org/problem?id=1182 (题目链接) 题意 中文题 Solution 带权并查集. 神犇博客,秒懂 fa记录父亲,r记录与父亲的关系.%3运用的很巧妙. 代码 / ...