PAT1028. List Sorting (25)---strcmp

题目链接为：https://www.patest.cn/contests/pat-a-practise/1028

1028. List Sorting (25)

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) 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

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

题目其实特别简单，应该是考察sort的使用。

<1>由于时间的限制，输入输出最好是scanf 和 printf，否则最后一个测试点将过不去;

<2>同样因为时间的限制，修改vector为数组（不过应该差别不会太大，主要还是<1>）。

论基础知识扎实的重要性，strcmp的结果是返回正数或者负数，而不是返回-1和1。这个点主要用在cmp函数中。

设这两个字符串为str1，str2，strcmp(str1,str2)的含义是：若str1=str2，则返回零；若str1<str2，则返回负数；若str1>str2，则返回正数。

 

以下是代码

 #include <bits/stdc++.h>
 using namespace std;
 #define maxn 100005
 #define For(I,A,B) for(int I = (A); I < (B); I++)

 class student
 {
     public:
     char id[],name[];
     int grade;
 };
 student stu[maxn];
 int n,c;
 bool cmp(const student &a,const student &b)
 {
     if(c == )
     {
         if(strcmp(a.id,b.id))
             return strcmp(a.id,b.id) < ;
     }
     else if(c == )
     {
         if(strcmp(a.name,b.name))
             return strcmp(a.name,b.name) < ;
         return strcmp(a.id,b.id) < ;
     }
     else if(c == )
     {
         if(a.grade != b.grade)
             return a.grade < b.grade;
         return strcmp(a.id,b.id) < ;
     }
 }

 int main()
 {
     freopen("1028.in","r",stdin);
     while(scanf("%d%d",&n,&c) != EOF)
     {
         //stu.clear();
         For(i,,n)
         {
             scanf("%s%s%d",&stu[i].id,&stu[i].name,&stu[i].grade);//>>stu[i].id>>stu[i].name>>stu[i].grade;
         }
         sort(stu,stu +n,cmp);
         For(i,,n)
         {
             printf("%s %s %d\n",stu[i].id,stu[i].name,stu[i].grade);//cout<<stu[i].id<<" "<<stu[i].name<<" "<<stu[i].grade<<endl;
         }
     }
     return ;
 }

PAT1028