poj 1458 Common Subsequence
来源:互联网 发布:网络致富平台 编辑:程序博客网 时间:2024/06/02 23:06
题意:赤裸裸的求最长公共序列。
算法思路:dp。
dp[i][j]指前i和j有多长的相同的个数。
代码:
- Common Subsequence--poj--1458
- poj 1458 Common Subsequence
- poj 1458 Common Subsequence
- Poj 1458 Common Subsequence
- POJ 1458 Common Subsequence
- poj 1458 Common Subsequence
- poj 1458 Common Subsequence
- POJ 1458 Common Subsequence
- poj 1458 Common Subsequence
- POJ-1458-Common Subsequence
- POJ 1458 Common Subsequence
- poj 1458 Common Subsequence
- POJ 1458 Common Subsequence
- poj 1458 Common Subsequence
- POJ 1458 Common Subsequence
- POJ 1458 Common Subsequence
- POJ 1458 Common Subsequence
- poj 1458 Common Subsequence
- 浅谈大脑思维与对未来计算机的设想
- 解决ThinkPad X200找不到无线网卡硬件的问题
- Servlet 3.0特性详解之可扩展性支持
- Servlet 3.0特性详解之 ServletContext 性能增强
- c++底层机制
- poj 1458 Common Subsequence
- 二叉树的前序、中序、后序遍历 查找 删除
- C++一维数组与指针
- 如何利用windosAPI计算程序运行时间 不使用VC的库,也就是说不跨系统,跨编译器
- Hibernate关系映射(1)_一对一单向外键关联
- 自己的心声
- poj 1519 Digital Roots
- 系统信息API
- boost ptime 与 time_t等的转换