uva 10192 Vacation

The Problem

You are planning to take some rest and to go out on vacation, but you really don’t know which cities you should visit. So, you ask your parents for help. Your mother says “My son, you MUST visit Paris, Madrid, Lisboa and London. But it’s only fun in this order.” Then your father says: “Son, if you’re planning to travel, go first to Paris, then to Lisboa, then to London and then, at last, go to Madrid. I know what I’m talking about.”

Now you’re a bit confused, as you didn’t expected this situation. You’re afraid that you’ll hurt your mother if you follow your father’s suggestion. But you’re also afraid to hurt your father if you follow you mother’s suggestion. But it can get worse, because you can hurt both of them if you simply ignore their suggestions!

Thus, you decide that you’ll try to follow their suggestions in the better way that you can. So, you realize that the “Paris-Lisboa-London” order is the one which better satisfies both your mother and your father. Afterwards you can say that you could not visit Madrid, even though you would’ve liked it very much.

If your father have suggested the “London-Paris-Lisboa-Madrid” order, then you would have two orders, “Paris-Lisboa” and “Paris-Madrid”, that would better satisfy both of your parent’s suggestions. In this case, you could only visit 2 cities.

You want to avoid problems like this one in the future. And what if their travel suggestions were bigger? Probably you would not find the better way very easy. So, you decided to write a program to help you in this task. You’ll represent each city by one character, using uppercase letters, lowercase letters, digits and the space. Thus, you can have at most 63 different cities to visit. But it’s possible that you’ll visit some city more than once.

If you represent Paris with “a”, Madrid with “b”, Lisboa with “c” and London with “d”, then your mother’s suggestion would be “abcd” and you father’s suggestion would be “acdb” (or “dacb”, in the second example).

The program will read two travel sequences and it must answer how many cities you can travel to such that you’ll satisfy both of your parents and it’s maximum.

The Input

The input will consist on an arbitrary number of city sequence pairs. The end of input occurs when the first sequence starts with an “#”character (without the quotes). Your program should not process this case. Each travel sequence will be on a line alone and will be formed by legal characters (as defined above). All travel sequences will appear in a single line and will have at most 100 cities.

The Output

For each sequence pair, you must print the following message in a line alone:

Case #d: you can visit at most K cities.

Where d stands for the test case number (starting from 1) and K is the maximum number of cities you can visit such that you’ll satisfy both you father’s suggestion and you mother’s suggestion.

Sample Input

abcd

acdb

abcd

dacb

#

Sample Output

Case #1: you can visit at most 3 cities.

Case #2: you can visit at most 2 cities.

题目大意:最长公共子序列。

解题思路:最长公共子序列。注意要用gets来读。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define N 150
using namespace std;
char a[N], b[N];
int dp[N][N];
int main() {
int Case = 1;
while (gets(a) != NULL) {
if (a[0] == '#') break;
gets(b);
memset(dp, 0, sizeof(dp));
int l1 = strlen(a), l2 = strlen(b);
int Max = 0;
for (int i = 1; i <= l1; i++) {
for (int j = 1; j <= l2; j++) {
if (a[i - 1] == b[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
}
else {
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
printf("Case #%d: you can visit at most %d cities.\n", Case++, dp[l1][l2]);
}
return 0;
}

版权声明:本文博主原创文章。博客不能未经同意转载。

uva 10192 Vacation(最长公共子)的更多相关文章

  1. uva 10066 The Twin Towers (最长公共子)

    uva 10066 The Twin Towers 标题效果:最长公共子. 解题思路:最长公共子. #include<stdio.h> #include<string.h> # ...

  2. 使用后缀数组寻找最长公共子字符串JavaScript版

    后缀数组很久很久以前就出现了,具体的概念读者自行搜索,小菜仅略知一二,不便讨论. 本文通过寻找两个字符串的最长公共子字符串,演示了后缀数组的经典应用. 首先需要说明,小菜实现的这个后缀数组算法,并非标 ...

  3. UVA.10192 Vacation (DP LCS)

    UVA.10192 Vacation (DP LCS) 题意分析 某人要指定旅游路线,父母分别给出了一系列城市的旅游顺序,求满足父母建议的最大的城市数量是多少. 对于父母的建议分别作为2个子串,对其做 ...

  4. UVa 10192 - Vacation &amp; UVa 10066 The Twin Towers ( LCS 最长公共子串)

    链接:UVa 10192 题意:给定两个字符串.求最长公共子串的长度 思路:这个是最长公共子串的直接应用 #include<stdio.h> #include<string.h> ...

  5. LIS(最长的序列)和LCS(最长公共子)总结

    LIS(最长递增子序列)和LCS(最长公共子序列)的总结 最长公共子序列(LCS):O(n^2) 两个for循环让两个字符串按位的匹配:i in range(1, len1) j in range(1 ...

  6. UVA 10192 Vacation

    裸最长公共子序列 #include<time.h> #include <cstdio> #include <iostream> #include<algori ...

  7. POJ 3356 AGTC(最长公共子)

    AGTC Description Let x and y be two strings over some finite alphabet A. We would like to transform  ...

  8. KMP该算法解释(最长公共子)

    一个:介绍KMP算法之前,首先解释一下BF算法 (1)BF算法(传统的匹配算法,是最简单的算法) BF算法是一种常见的模式匹配算法,BF该算法的思想是目标字符串S模式串的第一个字符P的第一个字符,以匹 ...

  9. POJ 2774 后缀数组:查找最长公共子

    思考:其实很easy.就在两个串在一起.通过一个特殊字符,中间分隔,然后找到后缀数组的最长的公共前缀.然后在两个不同的串,最长是最长的公共子串. 注意的是:用第一个字符串来推断是不是在同一个字符中,刚 ...

随机推荐

  1. 总结showModalDialog在开发中的一些问题

    一.在页面调用window.open()函数后,可以直接在打开的页面中用window.opener来调用父页面的方法,然而如果用showModalDialog打开一个模态窗口,就不能通过window. ...

  2. Codeforces 474B Worms 二分法(水

    主题链接:http://codeforces.com/contest/474/problem/B #include <iostream> #include <cmath> #i ...

  3. 一个Java对象到底占多大内存?(转)

    最近在读<深入理解Java虚拟机>,对Java对象的内存布局有了进一步的认识,于是脑子里自然而然就有一个很普通的问题,就是一个Java对象到底占用多大内存? 在网上搜到了一篇博客讲的非常好 ...

  4. JAVA WEB开发环境搭建教程

    一.下载安装JDK,配置好环境变量.(例如我JDK安装的目录为:C:\Program Files (x86)\Java\jdk1.6.0_10     ) 点击我的电脑-属性-系统设置(高级系统设置) ...

  5. C++学习笔记10-面向对象

    1.  面向对象的程序设计是基于三个基本概念:数据抽象.继承和动态绑定. 在C++ 在,凭借一流的数据抽象,随着一类从一个类派生还继承:派生类的成员继承基类.决定是使用基类中定义的函数还是派生类中定义 ...

  6. python 学习笔记 10 -- 正則表達式

    零.引言 在<Dive into Python>(深入python)中,第七章介绍正則表達式,开篇非常好的引出了正則表達式,以下借用一下:我们都知道python中字符串也有比較简单的方法, ...

  7. Extjs 3.4 和 web SSH(Ajaxterm)-howge-ChinaUnix博客

    Extjs 3.4 和 web SSH(Ajaxterm)-howge-ChinaUnix博客   Extjs 3.4 和 web SSH(Ajaxterm) 2013-04-07 15:20:17 ...

  8. 【C语言疯狂讲义】(八)C语言一维数组

    1.数组的基本概念: 同样类型    若干个     有序 由若干个同样类型的数据组成的有序的集合 有序:存储地址连续 下标连续 数组名:用来存放数组首地址的变量 数组元素:构成数组的每个数据 数组的 ...

  9. Android 程序静态分析

    简介 静态分析是探索Android程序内幕的一种最常见的方法,它与动态调剂双剑合璧,帮助分析人员解决分析时遇到的各种“疑难”问题. 静态分析是指在不运行的情况下,采用词法分析.语法分析等各种技术手段对 ...

  10. 前后端分离Web项目中,RBAC实现的研究

    在前后端分离Web项目中,RBAC实现的研究   最近手头公司的网站项目终于渐渐走出混沌,走上正轨,任务也轻松了一些,终于有时间整理和总结一下之前做的东西. 以往的项目一般使用模板引擎(如ejs)渲染 ...