uva 10420 List of Conquests（检索+排序）

List of Conquests
Input: standard input
Output: standard output
Time Limit: 2 seconds

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, il catalogo 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

3

Spain Donna Elvira

England Jane Doe

Spain Donna Anna

Sample Output

England 1

Spain 2

题目大意：给出n条语句，第一个单词为国家名2， 后面跟着一个女人的名字，表示这个女人属于这个国家，随后输出每个国家所拥有的女人数（按国家名的字典序）

解题思路：其实只要记录国家名出现的个数就可以了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define MAX 2005
#define N 205
#define M 80

struct country{
	char city[M];
	char women[N][M];
	int cnt;
	country(){
		memset(city, 0, sizeof(city));
		memset(women, 0, sizeof(women));
		cnt = 0;
	}
};

int cmp(const country &a, const country &b){
	return strcmp(a.city, b.city) < 0;
}

country man[MAX];
int main(){
	int n, leap = 0, bo;
	char str[N], name[M];

	// Read.
	scanf("%d", &n);
	for (int i = 0; i < n; i++){
		bo = 0;
		scanf("%s", man[leap].city);
		gets(man[leap].women[man[leap].cnt++]);

		for (int j = 0; j < leap; j++){
			if (strcmp(man[leap].city, man[j].city) == 0){
				bo = 1;
				strcpy(man[j].women[man[j].cnt++], man[leap].women[man[leap].cnt - 1]);
				break;
			}
		}
		if (bo)
			man[leap].cnt = 0;
		else
			leap++;
	}

	sort(man, man + leap, cmp);	

	for (int i = 0; i < leap; i++)
		printf("%s %d\n", man[i].city, man[i].cnt);
	return 0;
}