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；

#include <algorithm>
#include <stack>
#include <istream>
#include <stdio.h>
#include <map>
#include <math.h>
#include <vector>
#include <iostream>
#include <queue>
#include <string.h>
#include <set>
#include <cstdio>
#define FR(i,n) for(int i=0;i<n;i++)
#define MAX 2005
#define mkp pair <int,int>
using namespace std;
#include <bits/stdc++.h>
const int maxn = 5e5 + 40;
typedef long long ll;
const int  inf = 0x3fffff;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar());
    x *= 1 - 2 * flag;
}

//ll val[maxn],id[maxn];

struct inof{
   int id,val;
}p[maxn];
bool cmp(inof a,inof b)
{
    return a.val>b.val;
}

int res[maxn],top=0;
int Rank[maxn],fa[maxn];
int vis[maxn];


int fin(int x)
{
    if(x==fa[x])return x;
    else {
        return fa[x]=fin(fa[x]);
    }
}
int main() {
    int n;
    read(n);
    for(int i=1;i<=n;i++){
        read(p[i].val);
        p[i].id=i;
        fa[i]=i;
        Rank[i]=1;
    }
    sort(p+1,p+n+1,cmp);
    for(int i=1;i<=n;i++){
        int id=p[i].id;
        vis[id]=1;
        if(id!=n){
            if(vis[id+1]){
                int x=fin(id+1);
                fa[x]=id;
                Rank[id]+=Rank[x];
            }
        }
        if(id!=1){
            if(vis[id-1]){
                int x=fin(id-1);
                fa[x]=id;
                Rank[id]+=Rank[x];
            }
        }
        for(int j=top;j<Rank[id];j++)res[top++]=p[i].val;
       // top=Rank[id];
    }
    for(int i=0;i<top;i++)printf("%d ",res[i]);
    return 0;
}