1028 List Sorting (25分)
题目地址
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
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=100010;
struct node{
string id,name,score;
}s[maxn];
bool cmp1(node a,node b){
return a.id<b.id;
}
bool cmp2(node a,node b){
if(a.name!=b.name) return a.name<b.name;
else if(a.id!=b.id) return a.id<b.id;
}
bool cmp3(node a,node b){
if(a.score!=b.score) return a.score<b.score;
else if(a.id!=b.id) return a.id<b.id;
}
int main(){
int n,c;
cin>>n>>c;
for(int i=0;i<n;i++) cin>>s[i].id>>s[i].name>>s[i].score;
switch(c){
case 1:sort(s,s+n,cmp1);break;
case 2:sort(s,s+n,cmp2);break;
case 3:sort(s,s+n,cmp3);break;
}
for(int i=0;i<n;i++){
cout<<s[i].id<<" "<<s[i].name<<" "<<s[i].score<<endl;
}
}

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