【TJOI2010】中位数

Description

设计一种数据结构，支持元素的插入，查询当前元素的中位数

Solution

为了练习Splay，我决定用牛刀杀鸡，用Splay解决对顶堆的问题。

然后这就是一道Splay的模板题（简化简化再简化）

Code

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int INF = 2147483647;
 4 struct Splay {
 5     int sum, cnt, val;
 6     int fa, ch[2];
 7 } a[100010];
 8 int n, m, root, tot, cnt;
 9 inline int read() {
10     int ret = 0, op = 1;
11     char c = getchar();
12     while (!isdigit(c)) {
13         if (c == ‘-‘) op = -1;
14         c = getchar();
15     }
16     while (isdigit(c)) {
17         ret = ret * 10 + c - ‘0‘;
18         c = getchar();
19     }
20     return ret * op;
21 }
22 void update(int now) {
23     a[now].sum = a[a[now].ch[0]].sum + a[a[now].ch[1]].sum + a[now].cnt;
24 }
25 void connect(int x, int fa, int op) {
26     a[x].fa = fa;
27     a[fa].ch[op] = x;
28 }
29 void rotate(int x) {
30     int y = a[x].fa;
31     int z = a[y].fa;
32     int xson = a[y].ch[1] == x ? 1 : 0;
33     int yson = a[z].ch[1] == y ? 1 : 0;
34     int B = a[x].ch[xson ^ 1];
35     connect(B, y, xson); connect(y, x, xson ^ 1); connect(x, z, yson);
36     update(y); update(x);
37 }
38 void splay(int from, int to) {
39     while (a[from].fa != to) {
40         int y = a[from].fa;
41         int z = a[y].fa;
42         if (z != to)
43             (a[y].ch[0] == from) ^ (a[z].ch[0] == y) ? rotate(from) : rotate(y);
44         rotate(from);
45     }
46     if (to == 0) root = from;
47 }
48 void insert(int val) {
49     int now = root, fa = 0;
50     while (now && a[now].val != val) fa = now, now = a[now].ch[val > a[now].val];
51     if (now) a[now].cnt++;
52     else {
53         now = ++tot;
54         a[tot].val = val;
55         a[tot].fa = fa;
56         a[tot].sum = a[tot].cnt = 1;
57         a[tot].ch[0] = a[tot].ch[1] = 0;
58         if (fa) a[fa].ch[val > a[fa].val] = tot;
59     }
60     splay(now, 0);
61 }
62 int query(int k) {
63     int now = root;
64     while (1) {
65         if (k <= a[a[now].ch[0]].sum) now = a[now].ch[0];
66         else if (k > a[a[now].ch[0]].sum + a[now].cnt) {
67             k -= a[a[now].ch[0]].sum + a[now].cnt;
68             now = a[now].ch[1];
69         }
70         else return a[now].val;
71     }
72 }
73 int main() {
74     n = read();
75     insert(-INF); insert(INF);
76     for (register int i = 1; i <= n; ++i) ++cnt, insert(read());
77     m = read();
78     char s[10];
79     while (m--) {
80         scanf("%s", s);
81         if (s[0] == ‘a‘) ++cnt, insert(read());
82         else {
83             if (cnt & 1) printf("%d\n", query(cnt / 2 + 1 + 1));
84             else printf("%d\n", min(query(cnt / 2 + 1), query(cnt / 2 + 1 + 1)));
85         }
86     }
87     return 0;
88 }

AC Code

原文地址：https://www.cnblogs.com/shl-blog/p/11270173.html