1027. MJ, Nowhere to Hide

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

On BBS, there is a familiar term called MJ (short for MaJia), which means another BBS ID of one person besides his/her main ID.
These days, a lot of ACMers pour water on the ACMICPC Board of argo. Mr. Guo is very angry about that and he wants to punish these guys. ACMers are all smart boys/girls, right? They usually use their MJs while pouring water, so Mr. Guo can not tell all the IDs apart.  Unfortunately, the IP can not be changed, i.e, the posts of main ID and MJ of the same person has the same IP address, meanwhile, the IP addresses of different person is different.  Assuming that each person has exactly one main ID and one MJ, by reading their posts on BBS, you then tell Mr. Guo whom each MJ belongs to.

Input

The first line of each test cases is an even integer n (0<=n<=20), the number of posts on BBS.
Then n lines follow, each line consists of two strings:
BBS_ID IP_Address
BBS_ID means the ID who posts this post. BBS_ID is a string contains only lower case alphabetical characters and its length is not greater than 12. Each BBS ID appears only once in each test cases.
IP_Address is the IP address of that person. The IP address is formatted as “A.B.C.D”, where A, B, C, D are integers ranging from 0 to 255.
It is sure that there are exactly 2 different BBS IDs with the same IP address. The first ID appears in the input is the main ID while the other is the MJ of that person.
Your program should be terminated by n = 0.

Output

For each test case, output n/2 lines of the following format: “MJ_ID is the MaJia of main_ID”
They should be displayed in the lexicographical order of the main_ID.
Print a blank line after each test cases.
See the sample output for more details.

Sample Input

8
inkfish 192.168.29.24
zhi 192.168.29.235
magicpig 192.168.50.170
pegasus 192.168.29.235
iamcs 202.116.77.131
finalBob 192.168.29.24
tomek 202.116.77.131
magicduck 192.168.50.170
4
mmmmmm 172.16.72.126
kkkkkk 192.168.49.161
llllll 192.168.49.161
nnnnnn 172.16.72.126
0

Sample Output

tomek is the MaJia of iamcs
finalBob is the MaJia of inkfish
magicduck is the MaJia of magicpig
pegasus is the MaJia of zhi llllll is the MaJia of kkkkkk
nnnnnn is the MaJia of mmmmmm

Problem Source

ZSUACM Team Member

#include<iostream>
#include<map>
#include <string>
using namespace std; void swap(string&, string&);
void qsort(string [], int, int, string []);
int main() {
int n = 0;
while (cin>>n && n != 0) {
string id, ip;
map<string,string> acmers;
string former[n/2];
string latter[n/2];
int j = 0;
for (int i = 0; i < n; i++) {
cin >> id >> ip;
//使用map的支持小标为string型key,来达到直接字符串查找功能,
if (acmers.count(ip)==1) { //如果map中已经存在一个key为ip的项,则count为1,否则为0,从而判断出是否存在匹配
former[j] = id + " is the MaJia of ";
latter[j] = acmers[ip];
j++;
} else {
acmers[ip] = id;
}
}
qsort(latter, 0, j-1, former);
for (int i = 0; i < n/2; i++) {
cout << former[i] + latter[i] << endl;
}
cout << endl;
}
return 0;
}
//字符串交换
void swap(string &a, string &b) {
string temp = a;
a = b;
b = temp;
}
//快排实现字符串排序
void qsort(string latter[], int low, int height, string former[]) {
if (low < height) {
string lpvt = latter[(low+height)/2];
string fpvt = former[(low+height)/2];
int p = low;
swap(latter[(low+height)/2], latter[height]);
swap(former[(low+height)/2], former[height]);
for (int i = low; i < height; i++) {
if (latter[i].compare(0,latter[i].length(), lpvt) < 0) {
swap(latter[i], latter[p]);
swap(former[i], former[p]);
p++;
}
}
swap(latter[p], latter[height]);
swap(former[p], former[height]); qsort(latter, low, p-1, former);
qsort(latter, p+1, height, former);
}
}

  

sicily 1027 MJ, Nowhere to Hide 字符串匹配与排序的更多相关文章

  1. Sicily 1027. MJ, Nowhere to Hide

    //就是一个简单的字符串配对~~用map来解决很easy #include <iostream> #include <map> #include <string> ...

  2. 字符串匹配的KMP算法

    ~~~摘录 来源:阮一峰~~~ 字符串匹配是计算机的基本任务之一. 举例来说,有一个字符串”BBC ABCDAB ABCDABCDABDE”,我想知道,里面是否包含另一个字符串”ABCDABD”? 许 ...

  3. {Reship}{KMP字符串匹配}

    关于KMP字符串匹配的介绍和归纳,作者的思路非常清晰,推荐看一下 http://blog.csdn.net/v_july_v/article/details/7041827

  4. 字符串匹配(hash算法)

    hash函数对大家来说不陌生吧 ? 而这次我们就用hash函数来实现字符串匹配. 首先我们会想一下二进制数. 对于任意一个二进制数,我们将它化为10进制的数的方法如下(以二进制数1101101为例): ...

  5. 【C++实现python字符串函数库】二:字符串匹配函数startswith与endswith

    [C++实现python字符串函数库]字符串匹配函数startswith与endswith 这两个函数用于匹配字符串的开头或末尾,判断是否包含另一个字符串,它们返回bool值.startswith() ...

  6. sdut 2125串结构练习--字符串匹配【两种KMP算法】

    串结构练习——字符串匹配 Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目链接:http://acm.sdut.edu.cn/sduto ...

  7. C语言字符串匹配函数

    C语言字符串匹配函数,保存有需要时可以用: #include <stdio.h> #include <stdlib.h> #include <string.h> # ...

  8. 字符串匹配--Karp-Rabin算法

    主要特征 1.使用hash函数 2.预处理阶段时间复杂度O(m),常量空间 3.查找阶段时间复杂度O(mn) 4.期望运行时间:O(n+m) 本文地址:http://www.cnblogs.com/a ...

  9. 字符串匹配的KMP算法详解及C#实现

    字符串匹配是计算机的基本任务之一. 举例来说,有一个字符串"BBC ABCDAB ABCDABCDABDE",我想知道,里面是否包含另一个字符串"ABCDABD" ...

随机推荐

  1. 云服务器 ECS Linux IO 占用高问题排查方法

    https://help.aliyun.com/knowledge_detail/41224.html?spm=5176.7841174.2.19.uqC1as#使用 iostat 从系统纬度查看磁盘 ...

  2. 疑难杂症:NoSuchMethodError: com.opensymphony.xwork2.util.finder.UrlSet.includeClassesUrl(Lcom/opensymphony/xwork2/util/finder/ClassLoaderInterface;)

    严重: Exception starting filter struts2java.lang.NoSuchMethodError: com.opensymphony.xwork2.util.finde ...

  3. kmp算法简明教程

    在字符串s中寻找模式串p的位置,这是一个字符串匹配问题. 举例说明: i = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 s = a b a a c a b a a a b a a ...

  4. bing 输入法,切换简体、繁体快捷键与myeclipse 格式化代码冲突。。

    bing 输入法,切换简体.繁体快捷键与myeclipse 格式化代码冲突...蛋碎 myeclipse 代码格式化快捷键是:ctrl+shift+F,bing输入法简体.繁体切换也是,于是蛋疼的事情 ...

  5. 1.6.7 Detecting Languages During Indexing

    1. Detecting Languages During Indexing 在索引的时候,solr可以使用langid UpdateRequestProcessor来识别语言,然后映射文本到特定语言 ...

  6. NSTimer定时器的用法

    #import "ViewController.h" @interface ViewController () { NSTimer *countDownTimer; int cou ...

  7. Unity3D 使用脚本来控制 UI 的 Image 显示的图片。

    记录一下这个问题. 原文地址:http://tieba.baidu.com/p/3561719701 object obj = Resources.Load(资源名, typeof(Sprite)); ...

  8. [Java] JAVA程序员您需要学习的25个标准

    (1) 你需要精通面向对象分析与设计(OOA/OOD).涉及模式(GOF,J2EEDP)以及综合模式.你应该了解UML,尤其是class,object,interaction以 及statediagr ...

  9. 7. Android框架和工具之 android-percent-support-lib-sample(百分比支持)

    1. android-percent-support-lib-sample介绍: 谷歌最新的百分比布局库的示例项目.其实LinearLayout的layout_weight也能实现百分比效果,不过这个 ...

  10. iOS - 混合开发

    5.oc 与 js 混合开发 PhoneGap 专门做混合开发的,已经被另一个公司收购. WebViewJavascriptBridge 国内用的比较多 混合开发的好处:跨平台性.H5开发的时间成本, ...