【P2234 [HNOI2002]營業額統計】fhq_treap

P2234](https://www.luogu.org/problem/P2234)
其實就是問每一天的前驅和後繼 取個min即可 所以我們上平衡樹

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <ctime>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

inline int read()
{
    int date = 0,m = 1; char ch = 0;
    while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
    if(ch=='-'){m = -1; ch = getchar();}
    while(ch>='0' && ch<='9')
    {
        date = date*10+ch-'0';
        ch = getchar();
    }return date*m;
}

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
    #define mid ((l+r)>>1)
    struct node{int l,r,maxx;}T[MAX_N*40];

    #undef mid
}*/
const int MAX_N = 40025;
struct fhqtp
{
    int ch[2],rd,size,val;
}t[MAX_N];
int arr[MAX_N],root,r1,r2,r3,r4,cnt = 0;
namespace fhq
{
    void up(int rt) {t[rt].size = t[t[rt].ch[0]].size+t[t[rt].ch[1]].size+1;}
    void split(int x,int val,int &a,int &b)
    {
        if(!x) {a = b = 0;return;}
        if(t[x].val>val) b = x,split(t[x].ch[0],val,a,t[x].ch[0]);
        else a = x,split(t[x].ch[1],val,t[x].ch[1],b);
        up(x);
    }
    int Merge(int x,int y)
    {
        if(x==0||y==0) return x + y;
        if(t[x].rd<t[y].rd)
        {
            t[x].ch[1] = Merge(t[x].ch[1],y);
            up(x);return x;
        }
        else
        {
            t[y].ch[0] = Merge(x,t[y].ch[0]);
            up(y);return y;
        }
    }
    int newnode(int val)
    {
        t[++cnt].val = val;t[cnt].rd = rand();t[cnt].size = 1;
        return cnt;
    }
    void Insert(int val)
    {
        split(root,val,r1,r2);
        root = Merge(r1,Merge(newnode(val),r2));
    }
    int Kth(int rt,int k)
    {
        int x = rt;
        while(1)
        {
            int ls = t[x].ch[0],rs = t[x].ch[1];
            if(k<=t[ls].size) x = ls;
            else if(k>t[ls].size+1)
                k-=t[ls].size+1,x = rs;
            else return x;
        }
    }
    int pre(int val)
    {
        r1 = r2 = 0;split(root,val-1,r1,r2);
        int ans = t[Kth(r1,t[r1].size)].val;
        root = Merge(r1,r2);
        return ans;
    }
    int nex(int val)
    {
        r1 = r2 = 0;split(root,val,r1,r2);
        int ans = t[Kth(r2,1)].val;
        root = Merge(r1,r2);
        return ans;
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("turnover.in","r",stdin);
    //freopen("turnover.out","w",stdout);
    int n,ans = 0;srand(19260817);
    scanf("%d",&n);
    fhq::Insert(inf);fhq::Insert(_inf);
    for(int i = 1;i<=n;++i)
    {
        scanf("%d",&arr[i]);
        if(i==1) ans+=arr[i];
        else
        {
            ans+=min(abs(fhq::pre(arr[i]+1)-arr[i]),abs(fhq::nex(arr[i]-1)-arr[i]));
        }
        fhq::Insert(arr[i]);
    }
    printf("%d\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}