luogu P2947 [USACO09MAR]向右看齐Look Up |单调队列

题目描述

Farmer John's N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height H_i (1 <= H_i <= 1,000,000).

Each cow is looking to her left toward those with higher index numbers. We say that cow i 'looks up' to cow j if i < j and H_i < H_j. For each cow i, FJ would like to know the index of the first cow in line looked up to by cow i.

Note: about 50% of the test data will have N <= 1,000.

约翰的N(1≤N≤10^5)头奶牛站成一排，奶牛i的身高是Hi(l≤Hi≤1,000,000)．现在，每只奶牛都在向右看齐．对于奶牛i，如果奶牛j满足i<j且Hi<Hj，我们可以说奶牛i可以仰望奶牛j． 求出每只奶牛离她最近的仰望对象．

Input

输入格式

• Line 1: A single integer: N

• Lines 2..N+1: Line i+1 contains the single integer: H_i

第 1 行输入 N，之后每行输入一个身高 H_i。

输出格式

• Lines 1..N: Line i contains a single integer representing the smallest index of a cow up to which cow i looks. If no such cow exists, print 0.

共 N 行，按顺序每行输出一只奶牛的最近仰望对象，如果没有仰望对象，输出 0。

---

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=1e5+10;
int a[N],q[N],ans[N];
int main(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	scanf("%d",&a[i]);
	q[1]=1;
	int l=1,r=1;
	for(int i=1;i<=n;i++){
		while(r>=l&&a[q[r]]<a[i]){
			ans[q[r]]=i;
			r--;
		}
		q[++r]=i;
	}
	for(int i=1;i<=n;i++)printf("%d
",ans[i]);
}