1028 List Sorting（25 分）（C++)

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 (≤10​5​​) 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

解题思路：排序问题，按照题目要求即可。

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
using namespace std;
typedef struct{
  int id,score;
  string name;
}stu;
int cmp1(stu s1,stu s2){
  return s1.id<s2.id;
}
int cmp2(stu s1,stu s2){
  if(s1.name!=s2.name)
    return s1.name<s2.name;  
  else
    return s1.id<s2.id;
}
int cmp3(stu s1,stu s2){
  if(s1.score!=s2.score)
    return s1.score<s2.score;  
  else
    return s1.id<s2.id;
}
int main(){
  int n,m;
  cin>>n>>m;
  std::vector<stu> v(n);
  for(int i=0;i<n;i++){
    scanf("%d",&v[i].id);
    cin>>v[i].name;
    scanf("%d",&v[i].score);
  }
  switch(m){
    case 1:
        sort(v.begin(), v.end(),cmp1);
        break;
    case 2:
        sort(v.begin(), v.end(),cmp2);
        break;
    case 3:
        sort(v.begin(), v.end(),cmp3);
        break;
  }
  for(int i=0;i<v.size();++i){
    printf("%06d ",v[i].id);
    cout<<v[i].name;
    printf(" %d\n",v[i].score);
  }
  system("pause");
}