1028 List Sorting (25 分)
标签：ram exce otto ring first bit main HERE who
Excel can sort records according to any column. Now you are supposed to imitate this function.
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).
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.
3 1
000007 James 85
000010 Amy 90
000001 Zoe 60
000001 Zoe 60
000007 James 85
000010 Amy 90
4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98
000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60
4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90
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; }
PAT 甲级 1028 List Sorting (25 分)（排序，简单题）
标签：ram exce otto ring first bit main HERE who
原文地址：https://www.cnblogs.com/caiyishuai/p/11372233.html
