【PAT甲级 - 1028】List Sorting (25分)（模拟，排序）

题干：

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

 题目大意：

给定n个学生的学号，姓名，分数。让你按照一定的要求排序

解题报告：

按照题意模拟。注意sort中这种ifelse的，第二个别忘写return啊。。不然肯定就错了。然后，用cin会T。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
struct Node {
	int id;
	string name;
	int grd;
} R[MAX];
int n,op;
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 return a.id < b.id;
}
bool cmp3(Node a,Node b) {
	if(a.grd != b.grd) return a.grd < b.grd;
	else return a.id < b.id;
}
char s[MAX];
int main()
{
	cin>>n>>op;
	for(int i = 1; i<=n; i++) {
		scanf("%d%s%d",&R[i].id,s,&R[i].grd);
		R[i].name = s;
	}
	if(op == 1) sort(R+1,R+n+1,cmp1);
	if(op == 2) sort(R+1,R+n+1,cmp2);
	if(op == 3) sort(R+1,R+n+1,cmp3);
	for(int i = 1; i<=n; i++) {
		printf("%.6d %s %d\n",R[i].id,R[i].name.c_str(),R[i].grd);
	}


	return 0 ;
}