BZOJ 3224: Tyvj 1728 普通平衡树

3224: Tyvj 1728 普通平衡树

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 9629 Solved: 4091
[Submit][Status][Discuss]

Description

您需要写一种数据结构（可参考题目标题），来维护一些数，其中需要提供以下操作：
1. 插入x数
2. 删除x数(若有多个相同的数，因只删除一个)
3. 查询x数的排名(若有多个相同的数，因输出最小的排名)
4. 查询排名为x的数
5. 求x的前驱(前驱定义为小于x，且最大的数)
6. 求x的后继(后继定义为大于x，且最小的数)

Input

第一行为n，表示操作的个数,下面n行每行有两个数opt和x，opt表示操作的序号(1<=opt<=6)

Output

对于操作3,4,5,6每行输出一个数，表示对应答案

Sample Input

10
1 106465
4 1
1 317721
1 460929
1 644985
1 84185
1 89851
6 81968
1 492737
5 493598

Sample Output

106465
84185
492737

HINT

1.n的数据范围：n<=100000

2.每个数的数据范围：[-1e7,1e7]

数据如下http://pan.baidu.com/s/1jHMJwO2

Source

平衡树

[Submit][Status][Discuss]

FHQ Treap

  1 #include <bits/stdc++.h>
  2 
  3 const int N = 500005;
  4 
  5 int ls[N], rs[N], vl[N], tg[N], sz[N];
  6 
  7 inline int node(int v)
  8 {
  9     static int t = 1;
 10     sz[t] = 1;
 11     vl[t] = v;
 12     tg[t] = rand();
 13     return t++;
 14 }
 15 
 16 int merge(int a, int b)
 17 {
 18     if (!a || !b)return a + b;
 19     if (tg[a] > tg[b])
 20     {
 21         rs[a] = merge(rs[a], b);
 22         sz[a] = 1 + sz[ls[a]] + sz[rs[a]];
 23         return a;
 24     }
 25     else
 26     {
 27         ls[b] = merge(a, ls[b]);
 28         sz[b] = 1 + sz[ls[b]] + sz[rs[b]];
 29         return b;
 30     }
 31 }
 32 
 33 void split(int t, int k, int &a, int &b)
 34 {
 35     if (!t)a = b = 0;
 36     else
 37     {
 38         if (vl[t] <= k)
 39             a = t, split(rs[t], k, rs[t], b);
 40         else
 41             b = t, split(ls[t], k, a, ls[t]);
 42         sz[t] = 1 + sz[ls[t]] + sz[rs[t]];
 43     }
 44 }
 45 
 46 int kth(int t, int k)
 47 {
 48     if (k <= sz[ls[t]])
 49         return kth(ls[t], k);
 50     else if (k == sz[ls[t]] + 1)
 51         return t;
 52     else   
 53         return kth(rs[t], k - sz[ls[t]] - 1);
 54 }
 55 
 56 signed main(void)
 57 {
 58     srand(5264);
 59     
 60     int n, r = 0; scanf("%d", &n);
 61 
 62     for (int a, b, x, y, z; n--; )
 63     {
 64         scanf("%d%d", &a, &b);
 65 
 66         if (a == 1)
 67         {
 68             split(r, b, x, y);
 69             r = merge(x, node(b));
 70             r = merge(r, y);
 71         }
 72         else if (a == 2)
 73         {
 74             split(r, b, x, z);
 75             split(x, b - 1, x, y);
 76             y = merge(ls[y], rs[y]);
 77             r = merge(x, y);
 78             r = merge(r, z);
 79         }
 80         else if (a == 3)
 81         {
 82             split(r, b - 1, x, y);
 83             printf("%d\n", sz[x] + 1);
 84             r = merge(x, y);
 85         }
 86         else if (a == 4)
 87             printf("%d\n", vl[kth(r, b)]);
 88         else if (a == 5)
 89         {
 90             split(r, b - 1, x, y);
 91             printf("%d\n", vl[kth(x, sz[x])]);
 92             r = merge(x, y);
 93         }
 94         else
 95         {
 96             split(r, b, x, y);
 97             printf("%d\n", vl[kth(y, 1)]);
 98             r = merge(x, y);
 99         }
100     }
101 }

 1 #include <bits/stdc++.h>
 2 const int N = 500005;
 3 int ls[N], rs[N], vl[N], tg[N], sz[N], tot = 1;
 4 int node(int v) {
 5     return vl[tot] = v, sz[tot] = 1, tg[tot] = rand(), tot++;
 6 }
 7 int merge(int a, int b) {
 8     if (!a || !b)return a + b;
 9     if (tg[a] > tg[b]) {
10         rs[a] = merge(rs[a], b);
11         sz[a] = 1 + sz[ls[a]] + sz[rs[a]];
12         return a;
13     }
14     else {
15         ls[b] = merge(a, ls[b]);
16         sz[b] = 1 + sz[ls[b]] + sz[rs[b]];
17         return b;
18     }
19 }
20 int split(int t, int k, int &a, int &b) {
21     if (!t)return a = b = 0, 0;
22     if (vl[t] <= k)
23         a = t, split(rs[t], k, rs[t], b);
24     else
25         b = t, split(ls[t], k, a, ls[t]);
26     return sz[t] = 1 + sz[ls[t]] + sz[rs[t]];
27 }
28 int kth(int t, int k) {
29     return k <= sz[ls[t]] ? kth(ls[t], k) : ((k -= sz[ls[t]] + 1) ? kth(rs[t], k) : vl[t]);
30 }
31 signed main(void) {
32     int n, r = 0, a, b, x, y, z;
33     for (scanf("%d", &n); n--; ) {
34         scanf("%d%d", &a, &b);
35         if (a == 1)
36             split(r, b, x, y), r = merge(merge(x, node(b)), y);
37         else if (a == 2)
38             split(r, b, x, z), split(x, b - 1, x, y), r = merge(merge(x, merge(ls[y], rs[y])), z);
39         else if (a == 3)
40             split(r, b - 1, x, y), printf("%d\n", sz[x] + 1), r = merge(x, y);
41         else if (a == 4)
42             printf("%d\n", kth(r, b));
43         else if (a == 5)
44             split(r, b - 1, x, y), printf("%d\n", kth(x, sz[x])), r = merge(x, y);
45         else
46             split(r, b, x, y), printf("%d\n", kth(y, 1)), r = merge(x, y);
47     }
48 }