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 (≤) 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：
简单，三种排序可能，cmp写一下
AC代码:
#include<bits/stdc++.h>
using namespace std;
struct node
{
string num;
string name;
int score;
}a[100005];
bool cmp1(node x,node y){
return x.num<y.num;
};
bool cmp2(node x,node y){
if(x.name==y.name){
return x.num<y.num;
}
else return x.name<y.name;
};
bool cmp3(node x,node y){
if(x.score==y.score){
return x.num<y.num;
}
else return x.score<y.score;
};
int main()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++){
cin>>a[i].num>>a[i].name>>a[i].score;
}
if(m==1){
sort(a+1,a+1+n,cmp1);
}else if(m==2){
sort(a+1,a+1+n,cmp2);
}else{
sort(a+1,a+1+n,cmp3);
}
for(int i=1;i<=n;i++){
cout<<a[i].num<<" "<<a[i].name<<" "<<a[i].score<<endl;
}
return 0;
}
原文地址：https://www.cnblogs.com/caiyishuai/p/11372233.html