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.
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 }