http://acm.hdu.edu.cn/showproblem.php?pid=1908
题目大意是做一个操作系统，可执行4个系统操作，输入0（退出），1（输入P和K，并且以p排序后存储），2（输出p最小的那个k），3（输出p最大的那个k）注意每次输出后都要删掉那些输出的数据。。。。。我用一条双向链表写的，并设定了头结点和尾结点，，，这之中，头结点的next域存储了表头，尾结点的prior域存储了表尾，因为第一次写，，，，所以折磨了很久，每次输入P和K后都按p排序，然后一起存储
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1 #include<stdio.h> 2 #include<stdlib.h> 3 typedef struct node{ 4 int k; 5 int p; 6 struct node *next; 7 struct node *prior; 8 }S; 9 S *head,*last; 10 S *creat() 11 { 12 S *s; 13 s=(S *)malloc(sizeof(S)); 14 s->next=s->prior=NULL; 15 return s; 16 } 17 S *insert(int k,int p) 18 { 19 S *q=head->prior,*n=head->next,*r; 20 r=(S *)malloc(sizeof(S)); 21 r->k=k; 22 r->p=p; 23 if(n==last) 24 { 25 head->next=r; 26 r->next=last; 27 r->prior=head; 28 last->prior=r; 29 } 30 else { 31 while(n->next) 32 { 33 if(p<n->p&&n==head->next) 34 { 35 head->next=r; 36 r->next=n; 37 r->prior=head; 38 n->prior=r; 39 break; 40 } 41 if(p<n->p&&p>n->prior->p) 42 { 43 n->prior->next=r; 44 r->prior=n->prior; 45 r->next=n; 46 n->prior=r; 47 break; 48 } 49 if(p>n->p&&n==last->prior) 50 { 51 n->next=r; 52 r->prior=last->prior; 53 r->next=last; 54 last->prior=r; 55 break; 56 } 57 n=n->next; 58 } 59 } 60 } 61 void printmax() 62 { 63 S *s=head->next; 64 if(head->next==last) 65 printf("0\n"); 66 else 67 { 68 printf("%d\n",s->k); 69 s->next->prior=head; 70 head->next=s->next; 71 free(s); 72 } 73 } 74 S * printmin() 75 { 76 S *s=last->prior; 77 if(last->prior==head) 78 printf("0\n"); 79 else 80 { 81 printf("%d\n",s->k); 82 s->prior->next=last; 83 last->prior=s->prior; 84 free(s); 85 } 86 } 87 int main() 88 { 89 int n,k,p; 90 head=creat(); 91 last=creat(); 92 head->next=last; 93 last->prior=head; 94 while(scanf("%d",&n)&&n) 95 { 96 if(n==1) 97 { 98 scanf("%d%d",&k,&p); 99 insert(k,p); 100 } 101 else if(n==2) 102 printmin(); 103 else printmax(); 104 } 105 return 0; 106 }