洛谷 2234 BZOJ 1588 HNOI2002 营业额统计

Posted 2020-10-18 Driver_Lao

【题解】

　　treap模板题，直接用Treap维护前驱、后继，每次找到更接近当前val的加上就好了。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #define ls (a[u].l)
 4 #define rs (a[u].r)
 5 #define LL long long
 6 using namespace std;
 7 const int maxn=200010;
 8 int n,k,x,y,z,v,tot,root;
 9 struct treap{int l,r,v,rnd,size;}a[maxn];
10 LL sum=0;
11 inline void read(int &k){
12     k=0; int f=1; char c=getchar();
13     while(c<\'0\'||c>\'9\')c==\'-\'&&(f=-1),c=getchar();
14     while(\'0\'<=c&&c<=\'9\')k=k*10+c-\'0\',c=getchar();
15     k*=f;
16 }
17 void newnode(int v){a[++tot]=(treap){0,0,v,rand()+1,1};}
18 void update(int u){a[u].size=a[ls].size+a[rs].size+1;}
19 void split(int u,int k,int &x,int &y){
20     if(!k){x=0; y=u; return;}
21     if(a[u].size==k){x=u; y=0; return;}
22     if(a[ls].size>=k) split(ls,k,x,ls),y=u;
23     else split(rs,k-a[ls].size-1,rs,y),x=u;
24     update(u);
25 }
26 int merge(int x,int y){
27     if(!(x*y)) return x+y;
28     if(a[x].rnd>a[y].rnd){
29         a[y].l=merge(x,a[y].l); update(y); return y;
30     }
31     else{
32         a[x].r=merge(a[x].r,y); update(x); return x;
33     }
34 }
35 int qrank(int u,int val){
36     if(!u) return 0;
37     if(a[u].v>=val) return qrank(ls,val);
38     return qrank(rs,val)+a[ls].size+1;
39 }
40 int qval(int u,int k){
41     if(a[ls].size+1==k) return a[u].v;
42     return a[ls].size>=k?qval(ls,k):qval(rs,k-a[ls].size-1);
43 }
44 int main(){
45     srand(20020705); root=tot=1; a[root].v=2e9; a[root].size=1;
46     read(n);
47     for(int i=1;i<=n;i++){
48         read(v); 
49         if(i<=2) sum+=abs(v-sum);
50         else{
51             int tmp=qrank(root,v),ans=0X7f7f7f;
52             if(tmp) ans=abs(v-qval(root,tmp));
53             if(tmp<a[root].size) ans=min(ans,abs(v-qval(root,tmp+1)));
54             sum+=ans;
55         }
56         split(root,qrank(root,v),x,y);
57         newnode(v); root=merge(merge(x,tot),y);
58     }
59     return printf("%lld",sum),0;
60 }

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