CF91 B. Queue

题目传送门：https://codeforces.com/problemset/problem/91/B

题目大意：
给一个长度为\(n\)的序列\(A\)，记 \(B_i=j-i,j=\max\{j|i<j\and A_j<A_i\}\)，若不存在这样的 \(j\)，则记 \(B_i=-1\)，求\(B\)的值

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考虑用树状数组维护，倒序查询可以保证 \(i<j\) 这一限制条件，以权值作为下标，位置作为权值，便可以求出 \(A_j<A_i\) 中最大的 \(j\)

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
#define lowbit(x) ((x)&(-x))
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e5;
int A[N+10],Ans[N+10],list[N+10],Tree[N+10],n;
void Add(int x,int v){for (;x<=n;x+=lowbit(x))	Tree[x]=max(Tree[x],v);}
int Query(int x){
	int res=0;
	for (;x;x-=lowbit(x))	res=max(res,Tree[x]);
	return res;
}
int main(){
	n=read(0);
	for (int i=1;i<=n;i++)	A[i]=list[i]=read(0);
	sort(list+1,list+1+n);
	int T=unique(list+1,list+1+n)-list-1;
	for (int i=1;i<=n;i++)	A[i]=lower_bound(list+1,list+1+T,A[i])-list;
	for (int i=n;i>=1;i--){
		Ans[i]=Query(A[i]-1);
		Ans[i]=!Ans[i]?-1:Ans[i]-i-1;
		Add(A[i],i);
	}
	for (int i=1;i<=n;i++)	printf("%d%c",Ans[i],i==n?'\n':' ');
	return 0;
}