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
开始觉得本题25分太仁慈了，简直不要太简单。写完后发现超时。。。原来本题要用纯c代码写，否则最后一个数据太多cout,cin输入输出时间太长。
(纯c意味着不仅不能用cin/cout而且不能用string)纯c的函数还不是很熟悉.
要用char二维数组，纯c数组比较大小
strcmp(str1,str2) 字典序 str1<str2:-1 str1==str2:0 str1>str2:1 理解成str1-str2就行了
不能str1-str2 str1>str2 数组名是地址，比较的内存地址编号的大小，没有任何实际意义
#include<bits/stdc++.h>
using namespace std;
struct student{char info[4][10];//只是比大小 都设置成string没事
}stu[100010];
int N,C;
bool compare(student a,student b){if(strcmp(a.info[C],b.info[C])!=0) return strcmp(a.info[C],b.info[C])<0;//s1<s2:-1 s1==s2:0 s1>s2:1return strcmp(a.info[1],b.info[1])<0;
}
int main(){//freopen("in.txt","r",stdin);cin>>N>>C;for(int i=0;i<N;i++){for(int j=1;j<=3;j++) scanf("%s",stu[i].info[j]); }sort(stu,stu+N,compare);for(int i=0;i<N;i++){printf("%s %s %s\n",stu[i].info[1],stu[i].info[2],stu[i].info[3]);}return 0;
}