1028 List Sorting 排序

1028 List Sorting （25 分）

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

Input Specification:

Each input file contains one test case. For each case, the first line contains two integers N (≤105) 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).

Output Specification:

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.

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

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

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                

思路：
　　這個題與其說是考察排序，不如說是考察char數組和string之間的轉換，如果使用cin輸入會逾時

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<stdlib.h>
using namespace std;

struct Student
{
    string id;
    string name;
    int score;
};

bool cmp1(Student&A,Student&B)
{
    return A.id<B.id;
}

bool cmp2(Student&A,Student&B)
{
    return A.name==B.name?A.id<B.id:A.name<B.name;
}

bool cmp3(Student&A,Student&B)
{
    return A.score==B.score?A.id<B.id:A.score<B.score;
}


int main()
{
    int n,c;
    cin>>n>>c;
    Student stu[n];
    for(int i=0; i<n; i++)
    {
        char id[10],name[20];
        scanf("%s%s%d",id,name,&stu[i].score);
        stu[i].id=string(id);
        stu[i].name=string(name);

    }
    if(c==1)
    {
        //qsort(stu,n,sizeof(Student),cmp1);
        sort(stu,stu+n,cmp1);
    }
    else if(c==2)
    {
        sort(stu,stu+n,cmp2);
    }
    else
        sort(stu,stu+n,cmp3);
    for(int i=0; i<n; i++)
    {
        printf("%s %s %d\n",stu[i].id.c_str(),stu[i].name.c_str(),stu[i].score);
        //cout<<stu[i].id<<" "<<stu[i].name<<" "<<stu[i].score<<endl;

    }
    return 0;
}

轉載于:https://www.cnblogs.com/zhanghaijie/p/10295908.html