Election
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3558   Accepted: 1692

Description

Canada has a multi-party system of government. Each candidate is generally associated with a party, and the party whose candidates win the most ridings generally forms the government. Some candidates run as independents, meaning they are not associated with any party. Your job is to count the votes for a particular riding and to determine the party with which the winning candidate is associated.

Input

The first line of input contains a positive integer n satisfying 2 <= n <= 20, the number of candidates in the riding. n pairs of lines follow: the first line in each pair is the name of the candidate, up to 80 characters; the second line is the name of the party, up to 80 characters, or the word "independent" if the candidate has no party. No candidate name is repeated and no party name is repeated in the input. No lines contain leading or trailing blanks. 
The next line contains a positive integer m <= 10000, and is followed by m lines each indicating the name of a candidate for which a ballot is cast. Any names not in the list of candidates should be ignored. 

Output

Output consists of a single line containing one of:

  • The name of the party with whom the winning candidate is associated, if there is a winning candidate and that candidate is associated with a party.
  • The word "independent" if there is a winning candidate and that candidate is not associated with a party.
  • The word "tie" if there is no winner; that is, if no candidate receives more votes than every other candidate.

Sample Input

3
Marilyn Manson
Rhinoceros
Jane Doe
Family Coalition
John Smith
independent
6
John Smith
Marilyn Manson
Marilyn Manson
Jane Doe
John Smith
Marilyn Manson

Sample Output

Rhinoceros
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<iomanip>
#include<queue>
#include<map>
using namespace std; int main()
{
int Max = -;
bool flag = false;
map<string, string> ss;
map<string, int> st;
char Name[], Party[];
int n;
cin>>n;
getchar();
for (int i = ; i < n; i++)
{
gets(Name);
gets(Party);
ss[Name] = Party;
}
cin>>n;
getchar();
string result;
for (int i = ; i< n; i++)
{
gets(Name);
if (ss[Name] == "")
{
continue;
}
int nCount = ++st[Name];
if (nCount > Max)
{
flag = true;
result = Name;
Max = nCount;
}
else
{
if (nCount == Max)
{
flag = false;
}
}
}
if (flag)
{
cout<<ss[result]<<endl;
}
else
{
cout<<"tie"<<endl;
}
return ;
}

POJ 2643 Election的更多相关文章

  1. POJ 2643 Election map

    POJ 2643 Election 第一次写博客,想通过写博客记录自己的ACM历程,也想解释下英文题目,写些自己的理解.也可以让自己以后找题目更加方便点嘛.ElectionTime Limit: 10 ...

  2. POJ 2643

    #include<iostream> #include<stdio.h> #include<string> #include<algorithm> #d ...

  3. POJ 3664 Election Time 题解

    这道题网上非常多人都会说easy,水题之类的话,只是我看了下说这种话的人的程序,能够说他们的程序都不及格! 为什么呢?由于他们的程序都是使用简单的二次排序水过(大概你能搜索到的多是这种程序).那样自然 ...

  4. POJ 3905 Perfect Election(2-sat)

    POJ 3905 Perfect Election id=3905" target="_blank" style="">题目链接 思路:非常裸的 ...

  5. POJ 3905 Perfect Election (2-Sat)

    Perfect Election Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 438   Accepted: 223 De ...

  6. POJ 3905 Perfect Election

    2-SAT 裸题,搞之 #include<cstdio> #include<cstring> #include<cmath> #include<stack&g ...

  7. POJ 3905 Perfect Election (2-SAT 判断可行)

    题意:有N个人参加选举,有M个条件,每个条件给出:i和j竞选与否会只要满足二者中的一项即可.问有没有方案使M个条件都满足. 分析:读懂题目即可发现是2-SAT的问题.因为只要每个条件中满足2个中的一个 ...

  8. POJ 2528 Mayor's posters

    Mayor's posters Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Sub ...

  9. POJ 1625 Censored!(AC自动机+DP+高精度)

    Censored! Time Limit: 5000MS   Memory Limit: 10000K Total Submissions: 6956   Accepted: 1887 Descrip ...

随机推荐

  1. Java编程基础-面向对象(上)

    一.面向对象的概念 1.概念:面向对象是把解决的问题按照一定规则划分为多个独立的对象,然后通过调用对象的方法来解决问题.当然,一个应用程序会包含多个对象,通过多个对象的相互配合来实现应用程序的功能.这 ...

  2. Django中多表的增删改查操作及聚合查询、F、Q查询

    一.创建表 创建四个表:书籍,出版社,作者,作者详细信息 四个表之间关系:书籍和作者多对多,作者和作者详细信息一对一,出版社和书籍一对多 创建一对一的关系:OneToOne("要绑定关系的表 ...

  3. 学习用5W1H来管理自己的项目/工作

    学习用5W1H来管理自己的项目/工作   最近开始需要系统化的思维模型,这只是一个开始,一下用脑图的形式来简介5W1H的具体内容: 先写xmind思维树的文本导出,后面附上图片.^ _ ^ 5W1H ...

  4. SQLServer同一实例下事务操作

    参考代码: 引用Dapper public bool OrderAdd2(User user, Order order) { string dbString = ConfigurationManage ...

  5. iOS5 and iOS6都只支持横屏的方法

    If your app uses a UINavigationController, then you should subclass it and set the class in IB. You ...

  6. CPP-基础:函数指针,指针函数,指针数组

    函数指针 函数指针是指向函数的指针变量. 因而“函数指针”本身首先应是指针变量,只不过该指针变量指向函数.这正如用指针变量可指向整型变量.字符型.数组一样,这里是指向函数.如前所述,C在编译时,每一个 ...

  7. vs 2017 boost 安装目录 非安装

    linuxg++ -Wall -std=c++11 boost_socks5.cpp -o boost_socks5 -lboost_system -lboost_thread -lpthread m ...

  8. Java递归获取部门树 返回jstree数据

    @GetMapping("/getDept")@ResponseBodypublic Tree<DeptDO> getDept(String deptId){ Tree ...

  9. PAT (Basic Level) Practise (中文)-1033. 旧键盘打字(20)

    PAT (Basic Level) Practise (中文)-1033. 旧键盘打字(20)  http://www.patest.cn/contests/pat-b-practise/1033 旧 ...

  10. 关于jQuery中的$发生冲突及解决方案

    问题描述: 在Jquery库中,$是JQuery的别名,所有使用$的地方也都可以使用JQuery来替换,如$('#msg')等同于JQuery('#msg')的写法. 当引入多个js库后,其它的js库 ...