UVA - 10420 List of Conquests

最新推荐文章于 2019-02-16 23:05:35 发布

苏摩夜

最新推荐文章于 2019-02-16 23:05:35 发布

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分类专栏：排序和检索文章标签： uva UVA - 10420 List of

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本文介绍了一个算法挑战，该挑战要求读者帮助角色Leporello统计其主人唐乔凡尼在不同国家所追求女性的数量。输入包括一系列带有国籍和姓名的数据，输出则按字母顺序列出每个国家及相应的女性总数。

Description

In Act I, Leporello is telling Donna Elvira about his master's long list of conquests:

``This is the list of the beauties my master has loved, a list I've made out myself: take a look, read it with me. In Italy six hundred and forty, in Germany two hundred and thirty-one, a hundred in France, ninety-one in Turkey; but in Spain already a thousand and three! Among them are country girls, waiting-maids, city beauties; there are countesses, baronesses, marchionesses, princesses: women of every rank, of every size, of every age.'' (Madamina,ilcatalogo è questo)

As Leporello records all the ``beauties'' Don Giovanni ``loved'' in chronological order, it is very troublesome for him to present his master's conquest to others because he needs to count the number of ``beauties'' by their nationality each time. You are to help Leporello to count.

Input

The input consists of at most 2000 lines, but the first. The first line contains a number n, indicating that there will be n more lines. Each following line, with at most 75 characters, contains a country (the first word) and the name of a woman (the rest of the words in the line) Giovanni loved. You may assume that the name of all countries consist of only one word.

Output

The output consists of lines in alphabetical order. Each line starts with the name of a country, followed by the total number of women Giovanni loved in that country, separated by a space.

Sample Input

Spain Donna Elvira

England Jane Doe

Spain Donna Anna

Sample Output

England 1

Spain 2

字母的排序问题

<span style="font-family:Microsoft YaHei;font-size:18px;">#include<iostream>
#include<stdio.h>
#include<string>
#include<stdlib.h>
#include<sstream>
#include<algorithm>


using namespace std;

class Person
{
	public:
	string country;
	string name;


};

Person person[3000];

int comper(Person a,Person b){
	
	if(a.country == b.country)

		return a.name < b.name;
			
			else
				return a.country < b.country;


}

int main(){
	int n1;
	cin >> n1;
	string ck;
	getline(cin,ck);
	string str;
	for(int i = 0; i < n1; i ++){
		getline(cin,str);
			istringstream in(str);
		in >> person[i].country;

	while(!in.eof()){
		string t;
	    in >> t;
	    person[i].name += t;

	
	}	
	}
	sort(person,person + n1, comper);
	string m = person[0].country;
	string n = person[0].name;
	int count = 1;
	for(int i = 1;i < n1; i ++)
	{
		if(m == person[i].country)
		{
		 count ++ ;
		
		}
		else
		{
			cout << m << " " << count <<endl;
			m = person[i].country;
			n = person[i].name;
			count = 1;
		
		}
	
	}
	cout << m << " " << count << endl;
	return 0;
}

</span>