PAT (Advanced Level) Practice 1028 List Sorting (25分)

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 (≤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

自从上周考了冬季的乙级之后就再也没刷题了，这几天觉得心里有愧过来补补之前没做完的题目~

题意很清楚：

给定需要排序的条目总数，每一条分为三项：ID（由六位数构成，保证唯一），Name（可能重名），Score（可能重分）

当名字或者分数一样的情况下，选择ID的非降序排列，其余情况下针对输入操作数的要求进行某一特定项的非降序排列，比如输入1，就输出ID的非降序排列，其他以此类推~

水的不行，个人觉得只有乙级第三题难度...

#include
#define ll long long
#define rg register ll
#define maxn 100005
using namespace std;
struct node
{
    ll val,score;
    string name;
}p[maxn];
inline bool cmp1(const node&a,const node&b)
{
    return a.val>n>>opps;
    for(rg i=1;i<=n;i++)
    {
        cin>>p[i].val>>p[i].name>>p[i].score;
    }
    if(opps==1)sort(p+1,p+1+n,cmp1);
    if(opps==2)sort(p+1,p+1+n,cmp2);
    if(opps==3)sort(p+1,p+1+n,cmp3);
    for(rg i=1;i<=n;i++)
    {
        cout<