Codeforces Round #305 (Div. 2) D. Mike and Feet
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.
The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
10
1 2 3 4 5 4 3 2 1 6
6 4 4 3 3 2 2 1 1 1
学些数据结构才可以拯救自己吧,数据结构是省时的好东西哦,本来n^2可以变为nlgn甚至n,简直爆炸,单调栈就算能把复杂程度降到n的,太强辣,强行学习一波
其实我只是为了做出我们平台那道题,大佬讲让我用单调栈类似物,但是我并不了解这种数据结构啊,先mark下
#include <bits/stdc++.h> using namespace std; const int N=2e5+5; struct node { int L, num; }S; stack <node> sta; int a[N], ans[N]; int main () { int n; while (scanf ("%d", &n) != EOF) { for (int i=0; i<n; i++) scanf ("%d", &a[i]); for (int i=0; i<=n; i++) { int l = 0; while (!sta.empty ()) { S = sta.top (); if (S.num < a[i]) break; int lr = S.L + l; ans[lr] = max (ans[lr], S.num); l += S.L; sta.pop (); } sta.push((node){l+1, a[i]}); } for (int i=n; i>=1; i--) ans[i] = max (ans[i], ans[i+1]); for (int i=1; i<n; i++) printf ("%d ", ans[i]); printf ("%d\n", ans[n]); } return 0; }
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