BZOJ 1588 (treap)

题面

传送门

分析

语文题，主要是如何理解最小波动值

设当前天的营业额为x，则最小波动值为min(x-最大的<=x的数,最小的>=x的数-x)

然后用Treap维护序列就可以了

时间复杂度 $O(n \log n)$

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define lson tree[p].l
#define rson tree[p].r
#define INF 0x3f3f3f3f
#define maxn 100005 
using namespace std;
int n;
struct node{
	int l;
	int r;
	int cnt;
	int size;
	int val;
	int dat;
}tree[maxn];
int tot=0;
int root;
int New(int val){
	tree[++tot].cnt=1;
	tree[tot].size=1;
	tree[tot].val=val;
	tree[tot].dat=rand();
	return tot;
} 
 
void update(int p){
	tree[p].size=tree[lson].size+tree[rson].size+tree[p].cnt;
}
 
void build(){
	New(-INF);
	New(INF);
	root=1;
	tree[1].r=2;
	update(root);
}
 
void zig(int &p){
	int q=tree[p].l;
	tree[p].l=tree[q].r;
	tree[q].r=p;
	p=q;
	update(tree[p].r);
	update(p);
}
 
void zag(int &p){
	int q=tree[p].r;
	tree[p].r=tree[q].l;
	tree[q].l=p;
	p=q;
	update(tree[p].l);
	update(p);
}
 
void insert(int &p,int val){
	if(p==0){
		p=New(val);
		return;
	}
	if(tree[p].val==val){
		tree[p].cnt++;
		update(p);
		return;
	}
	if(val<tree[p].val){
		insert(lson,val);
		if(tree[p].dat<tree[lson].dat) zig(p);
	}else{
		insert(rson,val);
		if(tree[p].dat<tree[rson].dat) zag(p);
	}
	update(p);
}
 
int get_rank_by_val(int p,int val){
	if(p==0) return 0;
	if(tree[p].val==val) return tree[lson].size+1;
	else if(val<tree[p].val){
		return get_rank_by_val(lson,val);
	}else{
		return tree[lson].size+tree[p].cnt+get_rank_by_val(rson,val);
	}
}
 
int get_val_by_rank(int p,int rank){
	if(p==0) return INF;
	if(tree[lson].size>=rank) return get_val_by_rank(lson,rank);
	if(tree[lson].size+tree[p].cnt>=rank) return tree[p].val;
	return get_val_by_rank(rson,rank-tree[lson].size-tree[p].cnt); 
}
 
 
int main(){
	long long ans=0;
	int x,rk,low,up;
	build();
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&x);
		insert(root,x);
		if(i==1) ans+=x;
		else{
			rk=get_rank_by_val(root,x);
			low=get_val_by_rank(root,rk-1);
			up=get_val_by_rank(root,rk+1);
			ans+=min(abs(x-low),abs(x-up));
		}
	}
	printf("%lld\n",ans);
}