【PAT】1028. List Sorting (25)

题目描述

Excel can sort records according to any column. Now you are supposed to imitate this function.

翻译：Excel可以根据任何列对记录排序。现在你需要模仿这个机制。

INPUT FORMAT

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student’s record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).

翻译：一个输入文件包含一组测试数据。对于每组输入数据，第一行包括两个正整数N (<=100000) 和C，N代表记录数，C代表你需要根据第几行排序。接下来的N行，每行包括一个学生记录。一个学生记录包括他的唯一ID（一个六位数字），名字（一个不超过8位字符没有空格的字符串），和成绩（一个在0-100之间的整数）

OUTPUT FORMAT

For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID’s; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID’s in increasing order.

翻译：对于每组输入数据，输出N行排序结果。如果C=1，根据ID升序排序；如果C=2，根据姓名降序排序；如果C=3，记录必须根据成绩降序排序。如果有学生姓名相同或成绩相同，那么他们必须根据ID升序排序。

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Sample Input 1

3 1
000007 James 85
000010 Amy 90
000001 Zoe 60

Sample Output 1

000001 Zoe 60
000007 James 85
000010 Amy 90

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Sample Input 2

4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98

Sample Output 2

000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60

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Sample Input 3

4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90

Sample Output 3

000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90

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解题思路

定义一个结构体保存数据，重写三个排序函数，根据C来调用即可。读取数据需要用scanf，否则会超时。

#include
#include
#include
#include
#include
#include
#include
#include
#define INF 99999999
using namespace std;
struct Stu{
    int ID;
    string name;
    int grade;
    Stu(int id,string s,int g):ID(id),name(s),grade(g){}
};
bool cmp1(const Stu &a,const Stu &b){
    return a.IDbool cmp2(const Stu &a,const Stu &b){
    return a.name==b.name?a.IDbool cmp3(const Stu &a,const Stu &b){
    return a.grade==b.grade?a.IDvectorstu;
int N,C;
int main(){
    scanf("%d%d",&N,&C);
    string s;
    int id,grade;
    for(int i=0;iscanf("%d",&id);
        cin>>s;
        scanf("%d",&grade);
        stu.push_back(Stu(id,s,grade));
    }
    switch(C){
        case 1:sort(stu.begin(),stu.end(),cmp1);break;
        case 2:sort(stu.begin(),stu.end(),cmp2);break;
        case 3:sort(stu.begin(),stu.end(),cmp3);break;
    }
    for(int i=0;iprintf("%06d %s %d\n",stu[i].ID,stu[i].name.c_str(),stu[i].grade);
    }
    return 0;
}