sdut--Find a girl friend(结构体的二级排序)

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数据结构------排序

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本题描述了如何帮助CC找到班上年龄最小和最大的女生的名字。输入包含多个测试用例，每个用例首先给出女生数量，随后是各女生的年龄及名字。通过排序算法确定年龄最小和最大的女生，并输出其名字。

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Find a girl friend

Time Limit: 1000ms Memory limit: 262144K 有疑问？点这里^_^

题目描述

CC wants to find a girl friend, he knows each girl's age and name in his class, and each girl’s age is different..

Now he wants to know the youngest girl's name and the eldest girl’s name, can you help him?

输入

Input will consist of T test cases (1≤T≤10).

The first line of each test case contains a single integer n -- the number of girls (1≤n≤50).

Each of the following n lines contains Ai, Si, the ith girl's age and name (3≤age≤80, 2≤length(name)≤20), separated by exactly one space.

Each girl’s name consists of between 2 and 20 lowercase letters.

输出

For each test case, output the youngest girl's name and the eldest girl's name, seperated by exactly one space.

示例输入

5
21 liyuchun
18 fanbingbing
19 zhangziyi
22 liuyifei
20 liuyan
1
15 yangmi

示例输出

fanbingbing liuyifei
yangmi yangmi

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
using namespace std;
struct node
{
    int h;
    string n;
}q[10000];

bool cmp(node a,node b)
{
    return a.h<b.h;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            cin>>q[i].h>>q[i].n;
        }
        sort(q,q+n,cmp);
        cout<<q[0].n<<" "<<q[n-1].n<<endl;
    }
    return 0;
}