B. Mike and Feet Codeforces Round #305 (Div. 1) （并查集）

B. Mike and Feet

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.

蒟蒻不会单调队列的解法..等下好好研究下...

并查集的思路就是

先排序，由大到小，然后分别考虑每个位置的前后左右是否被联通，如果已经被联通，那么肯定比现在的数字要大；

 1 #include <algorithm>
 2 #include <stack>
 3 #include <istream>
 4 #include <stdio.h>
 5 #include <map>
 6 #include <math.h>
 7 #include <vector>
 8 #include <iostream>
 9 #include <queue>
10 #include <string.h>
11 #include <set>
12 #include <cstdio>
13 #define FR(i,n) for(int i=0;i<n;i++)
14 #define MAX 2005
15 #define mkp pair <int,int>
16 using namespace std;
17 #include <bits/stdc++.h>
18 const int maxn = 5e5 + 40;
19 typedef long long ll;
20 const int  inf = 0x3fffff;
21 void read(int &x) {
22     char ch; bool flag = 0;
23     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
24     for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar());
25     x *= 1 - 2 * flag;
26 }
27 
28 //ll val[maxn],id[maxn];
29 
30 struct inof{
31    int id,val;
32 }p[maxn];
33 bool cmp(inof a,inof b)
34 {
35     return a.val>b.val;
36 }
37 
38 int res[maxn],top=0;
39 int Rank[maxn],fa[maxn];
40 int vis[maxn];
41 
42 
43 int fin(int x)
44 {
45     if(x==fa[x])return x;
46     else {
47         return fa[x]=fin(fa[x]);
48     }
49 }
50 int main() {
51     int n;
52     read(n);
53     for(int i=1;i<=n;i++){
54         read(p[i].val);
55         p[i].id=i;
56         fa[i]=i;
57         Rank[i]=1;
58     }
59     sort(p+1,p+n+1,cmp);
60     for(int i=1;i<=n;i++){
61         int id=p[i].id;
62         vis[id]=1;
63         if(id!=n){
64             if(vis[id+1]){
65                 int x=fin(id+1);
66                 fa[x]=id;
67                 Rank[id]+=Rank[x];
68             }
69         }
70         if(id!=1){
71             if(vis[id-1]){
72                 int x=fin(id-1);
73                 fa[x]=id;
74                 Rank[id]+=Rank[x];
75             }
76         }
77         for(int j=top;j<Rank[id];j++)res[top++]=p[i].val;
78        // top=Rank[id];
79     }
80     for(int i=0;i<top;i++)printf("%d ",res[i]);
81     return 0;
82 }