CF91B：Queue（线段树，区间最值）

最新推荐文章于 2024-08-22 14:41:43 发布

转载最新推荐文章于 2024-08-22 14:41:43 发布 · 142 阅读

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原文链接：http://www.cnblogs.com/junior19/p/6729896.html

本文介绍了一种使用线段树解决特定队列问题的方法，该问题要求对于给定的一系列数字，计算每个元素右侧最远且小于当前元素的位置距离，若不存在则输出-1。通过构建线段树维护区间最小值并采用右子树优先查询策略，有效地解决了这一问题。

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There are n walruses standing in a queue in an airport. They are numbered starting from the queue's tail: the 1-st walrus stands at the end of the queue and the n-th walrus stands at the beginning of the queue. The i-th walrus has the age equal to ai.

The i-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such j (i < j), that ai > aj. The displeasure of the i-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the i-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.

The airport manager asked you to count for each of n walruses in the queue his displeasure.

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of walruses in the queue. The second line contains integers ai(1 ≤ ai ≤ 109).

Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.

Output

Print n numbers: if the i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the i-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.

Examples

input
6
10 8 5 3 50 45

output
2 1 0 -1 0 -1 

input
7
10 4 6 3 2 8 15

output
4 2 1 0 -1 -1 -1 

input
5
10 3 1 10 11

output
1 0 -1 -1 -1 

题意：给出N个数字，对于每个a[i]，输出在它右边最远的且比它小的数与它本身的距离，没有则输出-1。

思路：建个线段树保存着区间最小值，然后查询时右子树优先即可。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# define lson l, m, id<<1
# define rson m+1, r, id<<1|1
using namespace std;
const int maxn = 1e5;
int imin[maxn<<2], a[maxn+1], cnt=1;

void build(int l, int r, int id)
{
    if(l == r)
    {
        scanf("%d",&imin[id]);
        a[cnt++] = imin[id];
        return;
    }
    int m = (l+r)>>1;
    build(lson);
    build(rson);
    imin[id] = min(imin[id<<1], imin[id<<1|1]);
}

int query(int num ,int pos, int l, int r, int id)
{
    if(l == r)
        return a[l]<num?l:-1;
    int m = (l+r)>>1, ans=-1;
    if(imin[id<<1|1] < num)
        ans =  query(num, pos, rson);
    else if(m > pos && imin[id<<1] < num)//中位数比pos大，才执行左子树。
        ans = query(num, pos, lson);
    return ans;
}

int main()
{
    int n;
    scanf("%d",&n);
    build(1, n, 1);
    for(int i=1; i<=n; ++i)
    {
        int ans = query(a[i], i, 1, n, 1);
        printf("%d ",ans==-1?-1:ans-i-1);
    }
    return 0;
}

转载于:https://www.cnblogs.com/junior19/p/6729896.html