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 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, c, f, b, 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. 
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. 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
 
求最长公共子序列
 
 #include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;
int dp[][];
int main()
{
int lena,lenb;
char a[],b[];
while(scanf("%s %s",&a,&b)!=EOF)
{
int i,j;
lena=strlen(a);
lenb=strlen(b);
for(i = ; i <= lena ; i++)
{
for(j = ; j <= lenb ; j++)
{
if(a[i-] == b[j-])
dp[i][j]=dp[i-][j-]+;
else
dp[i][j]=max(dp[i][j-],dp[i-][j]);
}
}
printf("%d\n",dp[lena][lenb]);
}
}

杭电 1159 Common Subsequence的更多相关文章

  1. 杭电1159 Common Subsequence【最长公共子序列】

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 解题思路:任意先给出两个字符串 abcfbc abfcab,用dp[i][j]来记录当前最长的子 ...

  2. HDU 1159 Common Subsequence

    HDU 1159 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 = "b ...

  3. HDU 1159 Common Subsequence 公共子序列 DP 水题重温

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Jav ...

  4. hdu 1159 Common Subsequence(最长公共子序列)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Jav ...

  5. hdu 1159 Common Subsequence(最长公共子序列 DP)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Jav ...

  6. HDOJ 1159 Common Subsequence【DP】

    HDOJ 1159 Common Subsequence[DP] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K ...

  7. HDU 1159 Common Subsequence(裸LCS)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Jav ...

  8. HDU 1159 Common Subsequence 最长公共子序列

    HDU 1159 Common Subsequence 最长公共子序列 题意 给你两个字符串,求出这两个字符串的最长公共子序列,这里的子序列不一定是连续的,只要满足前后关系就可以. 解题思路 这个当然 ...

  9. 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 ...

随机推荐

  1. python入门之数据类型之列表、元组、字典

    list 格式: test_list = ["a",123,[1,"b"]] 索引: >>>print(test_list[0]) " ...

  2. bug 查找 (二) 从前端找到后端

    bug 查找 (二) 从前端找到后端 几天来,组长说我们系统的 apm 数据不正确,最体表现就是前端项目这几天错误统计为 0. 这不正常(没有办法,我们代码写的很烂),因为前端环境很复杂,网络,浏览器 ...

  3. Oracle查询排序asc/desc 多列 order by

    查询结果的排序 显示EMP表中不同的部门编号. 如果要在查询的同时排序显示结果,可以使用如下的语句: SELECT 字段列表 FROM 表名 WHERE 条件 ORDER BY 字段名1 [ASC|D ...

  4. 面向对象super内置函数(转)

    super函数用来解决钻石继承. 一.python的继承以及调用父类成员 父类: class Base(object): def __init__(self): print("base in ...

  5. [Ubuntu]清除系统磁盘垃圾

    操作步骤: 1.sudo apt-get autoremove(卸载系统中所有未被使用的依赖关系) 2.sudo apt-get clean(清除所有缓存的包文件) 以上操作绿色无害,对系统无影响.

  6. Spring Boot: Spring Starter Project

    好久没有创建过新项目,楼主发现Spring Boot项目创建失败了!!! 其中有两处错误: [图一不知道是哪里错,果断删掉重输入一次.成功进入下一步  其余步骤也没有错误,然而  最后一步失败了,如图 ...

  7. C# for循环的嵌套 作用域

    for() {   循环体可以套无数个for循环 } 比如:for() { for() { for() {... ...这里面可以镶嵌无数个for循环} } } 也可以这样 for() { for() ...

  8. LibreOJ #103. 子串查找

    题目描述 这是一道模板题. 给定一个字符串 A AA 和一个字符串 B BB,求 B BB 在 A AA 中的出现次数. A AA 中不同位置出现的 B BB 可重叠. 输入格式 输入共两行,分别是字 ...

  9. sourceTree配置bitbucket

    1. 为github增加账号信息 选择添加远程库 选择添加一个账号 输入用户名: 按照提示输入密码 选择bitbuchet为默认 选中搜索克隆

  10. 使用python模拟登陆百度

    #!/usr/bin/python # -*- coding: utf- -*- """ Function: Used to demostrate how to use ...