poj 1442 Treap实现名次树

Treap的入门题目,每个结点多维护一个size表示以它为根的子树的结点数,然后查kth的时候一层一层向下即可。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cstdio>
 5 #include <ctime>
 6 using namespace std;
 7 
 8 struct Node 
 9 {
10     Node * ch[2];
11     int r;      
12     int v;      
13     int s;
14     int cmp( int x )
15     {
16         if ( x == v ) return -1;
17         return x < v ? 0 : 1;
18     }    
19     void maintain()
20     {
21         s = 1;
22         if ( ch[0] != NULL ) s += ch[0]->s;
23         if ( ch[1] != NULL ) s += ch[1]->s;
24     }
25 };
26 
27 void rotate( Node * & o, int d )
28 {
29     Node * k = o->ch[d ^ 1];
30     o->ch[d ^ 1] = k->ch[d];
31     k->ch[d] = o;
32     o->maintain();
33     k->maintain();
34     o = k;
35 }
36 
37 void insert( Node * & o, int x )
38 {
39     if ( o == NULL )
40     {
41         o = new Node();
42         o->ch[0] = o->ch[1] = NULL;
43         o->v = x;
44         o->r = rand();
45         o->s = 1;
46     }
47     else
48     {
49         int d = ( x < o->v ? 0 : 1 );  
50         insert( o->ch[d], x );
51         if ( o->ch[d]->r > o->r )
52         {
53             rotate( o, d ^ 1 );
54         }
55         o->maintain();
56     }
57 }
58 
59 int kth( Node * o, int k )
60 {
61     int tmp = ( o->ch[0] == NULL ? 0 : o->ch[0]->s );
62     if ( k == tmp + 1 ) return o->v;
63     else if ( k < tmp + 1 ) return kth( o->ch[0], k );
64     else return kth( o->ch[1], k - tmp - 1 );
65 }
66 
67 const int N = 30001;
68 int a[N];
69 
70 int main ()
71 {
72     int n, m;
73     while ( scanf("%d%d", &n, &m) != EOF )
74     {
75         Node * root = NULL;
76         for ( int i = 1; i <= n; i++ )
77         {
78             scanf("%d", a + i);
79         }
80         int tt, cnt = 1;
81         for ( int i = 1; i <= m; i++ )
82         {
83             scanf("%d", &tt);
84             while ( cnt <= tt )
85             {
86                 insert( root, a[cnt++] );
87             }
88             printf("%d\n", kth( root, i ));
89         }
90     }
91     return 0;
92 }

 

posted @ 2015-08-02 10:26  hxy_has_been_used  阅读(253)  评论(0编辑  收藏  举报