# [HNOI2002] - Luogu 2234 - turnover statistics - tree

Keywords: C++

First exercise blog!!

This question is relatively simple.

The question is to input a sequence.

Every time you insert one, find the value that has the smallest difference with the previous one, and add their difference into the answer.

You can think of the balance tree here.

Every time you input it, add it to the tree, then find its precursor and successor, compare their difference, and add the small answer.

If you don't understand treap, you can read my previous blog: http://www.cnblogs.com/justin-cao/p/8270272.html

Post a code:

``` 1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<ctime>
6 using namespace std;
7 int n,root,size;
8 long long ans,sum;
9 long long inf=1000005;
10 long long a[50010];
11 struct P{
12     int l,r,sz,re,key,rd;
13 }t[50010];
14 void update(int k)
15 {
16     t[k].sz=t[t[k].l].sz+t[t[k].r].sz+t[k].re;
17 }
18 void left(int &k)
19 {
20     int y=t[k].r;
21     t[k].r=t[y].l;
22     t[y].l=k;
23     t[y].sz=t[k].sz;
24     update(k);
25     k=y;
26 }
27 void right(int &k)
28 {
29     int y=t[k].l;
30     t[k].l=t[y].r;
31     t[y].r=k;
32     t[y].sz=t[k].sz;
33     update(k);
34     k=y;
35 }
36 void init(int &k,int x)
37 {
38     if(k==0)
39     {
40         size++;
41         k=size;
42         t[k].sz=1;
43         t[k].re=1;
44         t[k].key=x;
45         t[k].rd=rand();
46         return;
47     }
48     t[k].sz++;
49     if(t[k].key==x)   t[k].re++;
50     else{
51         if(x>t[k].key)
52         {
53             init(t[k].r,x);
54             if(t[t[k].r].rd<t[k].rd)    left(k);
55         }
56         else{
57             init(t[k].l,x);
58             if(t[t[k].l].rd<t[k].rd)    right(k);
59         }
60     }
61 }
62 void pre(int k,int x)
63 {
64     if(k==0)  return;
65     if(t[k].key<=x)
66     {
67          ans=k;
68          pre(t[k].r,x);
69     }
70     else pre(t[k].l,x);
71 }
72 void nxt(int k,int x)
73 {
74     if(k==0)  return;
75     if(t[k].key>=x)
76     {
77         ans=k;
78         nxt(t[k].l,x);
79     }
80     else nxt(t[k].r,x);
81 }
82 int main()
83 {
84     srand(0);
85     scanf("%d",&n);
86     t[0].key=inf;
87     for(int i=1;i<=n;i++)  scanf("%lld",&a[i]);
88     init(root,a[1]);
89     for(int i=2;i<=n;i++)
90     {
91         pre(root,a[i]);
92         int x=ans;
93         nxt(root,a[i]);
94         sum+=(long long)min(abs(a[i]-t[x].key),abs(t[ans].key-a[i]));
95         init(root,a[i]);
96     }
97     printf("%lld",sum+a[1]);
98     return 0;
99 }```