HDU 1702 ACboy needs your help again!（单链表写法）

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can’t image how dark the room he was put into is, so poor ?.
As a smart ACMer, you want to get ACboy out of the monster’s labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can’t solve my problems, you will die with ACboy."
The problems of the monster is shown on the wall:
Each problem’s first line is a integer N(the number of commands), and a word “FIFO” or “FILO”.(you are very happy because you know “FIFO” stands for “First In First Out”, and “FILO” means “First In Last Out”).
and the following N lines, each line is “IN M” or “OUT”, (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!
Input
The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.
Output
For each command “OUT”, you should output a integer depend on the word is “FIFO” or “FILO”, or a word “None” if you don’t have any integer.
Sample Input

4
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT
5 FIFO
IN 1
IN 2
OUT
OUT
OUT
5 FILO
IN 1
IN 2
OUT
IN 3
OUT

Sample Output

1
2
2
1
1
2
None
2
3

题意描述： FILO先进后出，FIFO先进先出。

解题思路： 如果是FILO用为头插法建立链表，见到OUT就输出，如果是FIFO用尾插法建立链表，见到OUT就输出。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

struct node
{
	int data;
	struct node *next;
 } ;
 
 typedef struct node *LinkList;
 
 void wc(LinkList head,int n)//尾插法建立单链表
 {
 	LinkList p,rear=head;
 	char a[10];
 	int b;
 	for(int i=1;i<=n;i++)
 	{
 		scanf("%s",a);
 		if(a[0]=='I')
 		{
		 	p=(node *)malloc(sizeof(node));
 			scanf("%d",&b); 
 			p->data=b;
 			rear->next=p;
 			rear=p; 
		}
		rear->next=NULL;
		p=head->next;
		if(a[0]=='O')
		{
			p=head->next;
			if(p==NULL)
				printf("None\n");
			else
			{
				printf("%d\n",p->data);
				head=head->next;
			}
			
		}
	}
 }
 
 void tc(LinkList head,int n)//头插法建立单链表
 {
 	LinkList p;
 	char a[10];
 	int b;
	for(int i=1;i<=n;i++)
	{
		scanf("%s",a);
		if(a[0]=='I')
		{
			p=(node *)malloc(sizeof(node));
			scanf("%d",&b);
			p->data=b;
			p->next=head->next;
			head->next=p;
		}
		if(a[0]=='O')
		{
			p=head->next;
			if(p==NULL)
				printf("None\n");
			else
			{
				printf("%d\n",p->data);
				head=head->next;
			}
		
		}
	}
 }
 
 int main(void)
 {
 	LinkList head;
 	int n; 
 	char a[110],b[10];
 	int t;
 	scanf("%d",&t);
	 while(t--)
 	{
	 	scanf("%d",&n);
	 	getchar();
 		scanf("%s",b);
		if(b[2]=='F')
		{
			head=(node *)malloc(sizeof(node));
			head->next=NULL;
			wc(head,n);
		}	
		if(b[2]=='L')
		{
			head=(node *)malloc(sizeof(node));
			head->next=NULL;
			tc(head,n);	
		}	
	}
	return 0;
 }