NC24840 [USACO 2009 Mar S]Look Up

题目

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 Hi (1 <= Hi <= 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 Hi < Hj. 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.

输入描述

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

输出描述

• 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.

示例1

输入

6 
3 
2 
6 
1 
1 
2 

输出

3
3
0
6
6
0

说明

FJ has six cows of heights 3, 2, 6, 1, 1, and 2.
Cows 1 and 2 both look up to cow 3; cows 4 and 5 both look up to cow 6; and cows 3 and 6 do not look up to any cow.

题解

知识点：单调栈。

此题需要找到最邻近大于的牛，所以用单调栈维护即可。

单调栈和单调队列属于同一类原理，不同的是单调队列多了维护变区间的队头弹出，而单调栈不需要。这类单调结构除去单调队列里说过的区间最值的性质，还有一个查找最邻近大于/小于的功能，因为在元素与最邻近大于/小于之间的元素都是比自己小/大的，一定能被弹出，而之前最邻近大于/小于的元素在这之间也不会被弹出，所以新元素进入之前不被弹出的第一个元素即最邻近大于/小于的元素。

这里由于是要找右侧的，所以从右往左应用单调递减队列查找最邻近大于的元素。要注意如果栈空说明右侧没有比自己大的。

时间复杂度 $O(n)$

空间复杂度 $O(n)$

代码

#include <bits/stdc++.h>
 
using namespace std;
 
int h[100007];
 
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    for (int i = 1;i <= n;i++) cin >> h[i];
    stack<int> q;
    vector<int> ans;
    for (int i = n;i >= 1;i--) {
        while (!q.empty() && h[q.top()] <= h[i]) q.pop();
        ans.push_back(q.empty() ? 0 : q.top());
        q.push(i);
    }
    for (int i = n - 1;i >= 0;i--) cout << ans[i] << '\n';
    return 0;
}