【treap】6145 - Version Controlled IDE 2012合艾赛区E题

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4156

     比赛当时，我们队就一直坑在这个题上了，没有人看到最后一小段，一直用可以离线的splay企图AC，结果当然很惨烈=。=，耗费了半场比赛在这么一个不存再的题目上面。。。

     题目本质是一个普通的平衡树，但是限定了只能在线，所以写一个普通的函数式treap即可通过，参考了范浩强神牛的讲义，突然发现这个东西可以特别好写比普通的还短，具体看代码吧。。思路啥的仔细看代码不难懂。

View Code 
 1 //By Lin
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 
 7 struct    Node{
 8     int key,weight,size;
 9     Node *l,*r;
10     Node(int _key , int _weight, Node *_l, Node* _r):
11         key(_key),weight(_weight),l(_l),r(_r){
12             size = 1; 
13             if ( l ) size += l->size;
14             if ( r ) size += r->size;
15         }
16     Node *newnode(int key){
17         return new Node(key,rand(),NULL,NULL);
18     }
19     inline int lsize(){ return l?l->size:0; }
20     inline int rsize(){ return r?r->size:0; }
21 }*root[50005];
22 Node*    Meger(Node *a , Node *b ){
23     if ( !a || !b ) return a?a:b;
24     return a->weight>b->weight?
25          new Node(a->key,a->weight,a->l,Meger(a->r,b)):
26          new Node(b->key,b->weight,Meger(a,b->l),b->r);
27 }
28 Node*    Split_L(Node *a ,int size ){
29     if ( !a || size == 0 ) return NULL;
30     return a->lsize() < size?
31         new Node(a->key,a->weight,a->l,Split_L(a->r,size-1-a->lsize())):
32         Split_L(a->l,size);
33 }
34 Node*    Split_R(Node *a ,int size ){
35     if ( !a || size == 0 ) return NULL;
36     return a->rsize() < size?
37         new Node(a->key,a->weight,Split_R(a->l,size-1-a->rsize()),a->r):
38         Split_R(a->r,size);
39 }
40 int        Ask( Node *a ,int k ){
41     if ( a->lsize() >= k ) return Ask(a->l,k);
42     k -= a->lsize()+1;
43     if ( k == 0 ) return a->key;
44     return Ask(a->r,k);
45 }
46 int        len = 0;
47 
48 int        main(){
49     int d = 0 , cas;
50     scanf("%d", &cas );
51     root[0] = NULL;
52     int cnt = 1,kind,v,p,c;
53     char s[1005];
54     while ( cas -- ) {
55         scanf("%d", &kind );
56         if ( kind == 1 ) {
57             scanf("%d%s", &p , s );
58             p-=d;
59             Node *l = Split_L(root[cnt-1],p),
60                  *r = Split_R(root[cnt-1],len-p);
61             for (int i = 0; s[i]; i++ ){
62                 l = Meger(l,new Node(s[i],rand(),NULL,NULL));
63                 len++;
64             }
65             root[cnt++] = Meger(l,r);
66         }
67         else if ( kind == 2){
68             scanf("%d%d", &p , &c );
69             p-=d,c-=d;
70             Node *l = Split_L(root[cnt-1],p-1),
71                  *r = Split_R(root[cnt-1],len-p-c+1);
72             len -= c;
73             root[cnt++] = Meger(l,r);
74         }
75         else{
76             scanf("%d%d%d", &v, &p , &c );
77             v-=d,p-=d,c-=d;
78             char ch; 
79             for (int i = p; i<p+c; i++) {
80                 printf("%c", ch = Ask(root[v],i) );
81                 if ( ch == 'c' ) d++;
82             }
83             puts("");
84         }
85     }
86     return 0;
87 }
posted @ 2012-12-03 23:11  lzqxh  阅读(307)  评论(0)  编辑  收藏  举报