Codeforces Round #358 (Div. 2) D. Alyona and Strings 字符串dp
题目链接:
题目
D. Alyona and Strings
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
问题描述
After returned from forest, Alyona started reading a book. She noticed strings s and t, lengths of which are n and m respectively. As usual, reading bored Alyona and she decided to pay her attention to strings s and t, which she considered very similar.
Alyona has her favourite positive integer k and because she is too small, k does not exceed 10. The girl wants now to choose k disjoint non-empty substrings of string s such that these strings appear as disjoint substrings of string t and in the same order as they do in string s. She is also interested in that their length is maximum possible among all variants.
Formally, Alyona wants to find a sequence of k non-empty strings p1, p2, p3, ..., pk satisfying following conditions:
s can be represented as concatenation a1p1a2p2... akpkak + 1, where a1, a2, ..., ak + 1 is a sequence of arbitrary strings (some of them may be possibly empty);
t can be represented as concatenation b1p1b2p2... bkpkbk + 1, where b1, b2, ..., bk + 1 is a sequence of arbitrary strings (some of them may be possibly empty);
sum of the lengths of strings in sequence is maximum possible.
Please help Alyona solve this complicated problem and find at least the sum of the lengths of the strings in a desired sequence.
A substring of a string is a subsequence of consecutive characters of the string.
输入
In the first line of the input three integers n, m, k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10) are given — the length of the string s, the length of the string t and Alyona's favourite number respectively.
The second line of the input contains string s, consisting of lowercase English letters.
The third line of the input contains string t, consisting of lowercase English letters.
输出
In the only line print the only non-negative integer — the sum of the lengths of the strings in a desired sequence.
It is guaranteed, that at least one desired sequence exists.
样例
input
9 12 4
bbaaababb
abbbabbaaaba
output
7
题意
求由两个字符串的k个公共子串按顺序拼成的最长子序列。
题解
dp[i][j][kk][0]表示串s1[0...i]和s2[0...j]的kk个公共子串能拼的最长长度,并且最后一个字串还能继续连下去。
dp[i][j][kk][1]表示串s1[0...i]和s2[0...j]的kk个公共子串能拼的最长长度,并且最后一个字串不能继续连下去。
代码
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 1111;
typedef __int64 LL;
LL dp[maxn][maxn][11][2];
char s1[maxn], s2[maxn];
int n, m, k;
int main() {
scanf("%d%d%d", &n, &m, &k);
scanf("%s%s", s1+1, s2+1);
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
for (int kk = 1; kk <= k; kk++) {
if (s1[i] == s2[j]) {
dp[i][j][kk][0] = max(dp[i - 1][j - 1][kk][0], dp[i - 1][j - 1][kk - 1][1]) + 1;
}
dp[i][j][kk][1] = max(dp[i - 1][j][kk][1], dp[i][j - 1][kk][1]);
dp[i][j][kk][1] = max(dp[i][j][kk][1], dp[i][j][kk][0]);
}
}
}
printf("%I64d\n", dp[n][m][k][1]);
return 0;
}
Codeforces Round #358 (Div. 2) D. Alyona and Strings 字符串dp的更多相关文章
- Codeforces Round #358 (Div. 2) D. Alyona and Strings dp
D. Alyona and Strings 题目连接: http://www.codeforces.com/contest/682/problem/D Description After return ...
- Codeforces Round #358 (Div. 2) E. Alyona and Triangles 随机化
E. Alyona and Triangles 题目连接: http://codeforces.com/contest/682/problem/E Description You are given ...
- Codeforces Round #358 (Div. 2) C. Alyona and the Tree 水题
C. Alyona and the Tree 题目连接: http://www.codeforces.com/contest/682/problem/C Description Alyona deci ...
- Codeforces Round #358 (Div. 2) A. Alyona and Numbers 水题
A. Alyona and Numbers 题目连接: http://www.codeforces.com/contest/682/problem/A Description After finish ...
- Codeforces Round #358 (Div. 2)B. Alyona and Mex
B. Alyona and Mex time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Codeforces Round #358 (Div. 2) C. Alyona and the Tree dfs
C. Alyona and the Tree time limit per test 1 second memory limit per test 256 megabytes input standa ...
- Codeforces Round #358 (Div. 2)——C. Alyona and the Tree(树的DFS+逆向思维)
C. Alyona and the Tree time limit per test 1 second memory limit per test 256 megabytes input standa ...
- Codeforces Round #358 (Div. 2) C. Alyona and the Tree
C. Alyona and the Tree time limit per test 1 second memory limit per test 256 megabytes input standa ...
- 水题 Codeforces Round #302 (Div. 2) A Set of Strings
题目传送门 /* 题意:一个字符串分割成k段,每段开头字母不相同 水题:记录每个字母出现的次数,每一次分割把首字母的次数降为0,最后一段直接全部输出 */ #include <cstdio> ...
随机推荐
- JavaScript之返回顶部
为了弄这个Hexo,今天又是坐在电脑面前待了一天( ⊙ o ⊙ ),老是出问题,在百度上也试验了很多方法,还是没弄好,诶...身心疲惫甚是乏累啊~~~ 算了,这个Hexo先不弄,还是安分点吧,在Hex ...
- 计算日期时间 自动加1天 PHP计算闰年 java与PHP时间戳对比区别
昨天写一个同步数据库的模块 从一个数据库同步到另外一个数据库,因为数据较多,不可能一次性全部搬迁过去,所以就按照每天搬迁! 写了一个 模块,点击加1,只要点击一次,自动从A数据库取出1天的数据, 并 ...
- SQL取某个字段最大(小)数值及其相应行的其他字段值的句语
如下表Z 中,取 字段a 最大的那行 字段a 字段a 字段cSP000016964 5 20SP000016964 7 30SP000016964 1 15SP0000177 ...
- SQL Server 事务处理 回滚事务
--创建表: GO CREATE TABLE [dbo].[tb1]( [Id] [int] NOT NULL, [c1] [nvarchar](50) NULL, [c2] [datetime] N ...
- 如何系统地学习Node.js?
转载自知乎:http://www.zhihu.com/question/21567720 ------------------------------------------------------- ...
- JavaScript开发技巧
1.在编写js代码时,应尽量避免全局变量的使用.如果实在需要使用全局变量,则可以使用一个function来规避全局变量的使用. 2.数字解析 //1.丢弃小数部分,保留整数部分 alert( pars ...
- 使用Linux系统中的SSH服务
使用Linux系统中的SSH服务 1.SSH服务应用场景 ① 可以实现对文件的上传与下载 ② 实现远程管理Linux 2.安装SSH服务器 服 务:sshd 位 置:光盘2 软 件:openssh-s ...
- 使用单用户模式破解Linux密码
使用单用户模式破解Linux密码 特别说明:在实际工作应用中,安装Linux操作系统必须设置装载口令,否则很容易被破解. 1.使用reboot指令重启Linux操作系统 2.在进入操作系统数秒时,单击 ...
- spring IOC经典理解
不多解释,直接上图片!
- Windows 命令大全
打开控制面板的方法:输入control,回车即可打开. 以下是“运行”里常见的命令: gpedit.msc-----组策略 sndrec32-------录音机 Nslookup-------IP地址 ...