7-2 keep the linked list in order

For the information of several students entered , Establish an orderly linked list from small to large according to the student number , Finally, traverse the linked list , And output student information in order .

Input format :

First enter a positive integer T, Number of groups representing test data , And then there was T Group test data . First, input a positive integer for each group of test data n（1≤n≤100）, Number of students . Then input n Line information , They are the student number and name of the student , among , The student number is 8 Bit positive integer （ Guarantees vary ）, The name is no longer than 10 A string without spaces .

Output format :

For each group of tests , Output student information in order , Leave a space between student number and name （ See the output example ）.

Input sample mile :

1
3
20080108 Zhangsan
20070328 Lisi
20070333 Wangwu

sample output :

20070328 Lisi
20070333 Wangwu
20080108 Zhangsan

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct stu{
	int num;
	char name[100];
	struct stu *next;
};
int main(){
	int n=0,m,i=0,j=0;
	struct stu *head=NULL,*p,*tail,*q,*index,temp;
	scanf("%d",&n);
	while(i<n){
		scanf("%d",&m);
		for(j=0;j<m;j++){
			
			p=(struct stu *)malloc(sizeof(struct stu ));
		    scanf("%d %s\n",&p->num,p->name);
		    if(head==NULL)  head=tail=p;
		    else{
		    	tail->next=p;
			}
			tail=p;
		}
		tail->next=NULL;
		i++;
	}
	for(p=head;p!=NULL;p=p->next){
		for(q=p->next;q!=NULL;q=q->next){
			if(q->num<p->num){
				temp.num=p->num;
				strcpy(temp.name,p->name);
				
				p->num=q->num;
				strcpy(p->name,q->name);
				
				q->num=temp.num;
				strcpy(q->name,temp.name);
			}
		}	
	}
	for(p=head;p!=NULL;p=p->next){
		printf("%d %s\n",p->num,p->name);
	}
	return 0;
}