UVa 10420 List of Conquests

题意就是有N个pl妹子，然后每行第一个单词是妹子的国籍，后面是妹子的名字。

你的任务就是统计相同国籍妹子的个数，然后按字母表顺序输出。

我首先把所有的国籍都读入，然后用qsort()按字母表顺序排序。

 

 

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

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

``This is the list of the beauties my master has loved, a list I've madeout myself: take a look, read it with me. In Italy six hundred and forty, inGermany 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 hismaster's conquest to others because he needs to count the number of``beauties'' by their nationality each time. You are to help Leporello tocount.

Input

The input consists of at most 2000 lines, but the first. The first linecontains a number n,indicating that there will be n more lines. Each following line, withat most 75 characters, contains a country (thefirst 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 onlyone word.

Output

The output consists of lines in alphabetical order. Eachline starts with the name of a country, followed by the total number of womenGiovanni 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

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Problem-setter: Thomas Tang,Queens University, Canada

 

AC代码：

//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int main(void)
{
    #ifdef LOCAL
        freopen("10420in.txt", "r", stdin);
    #endif

    int cmp(const void *a, const void *b);
    int N, i, j;
    char country[2005][80], c[200];
    scanf("%d", &N);
    for(i = 0; i < N; ++i)
    {
        scanf("%s", country[i]);
        gets(c);
    }
    qsort(country, N, sizeof(country[0]), cmp);

    i = 0;
    while(i < N)
    {
        j = i;
        while(strcmp(country[j], country[i]) == 0 && j < N)
        {
            ++j;
        }
        cout << country[i] << " " << j - i << endl;
        i = j;
    }
    return 0;
}
int cmp(const void *a, const void *b)
{
    return strcmp((char *)a, (char *)b);
}