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Description
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.
Sample Input
10
1 2 3 4 5 4 3 2 1 6
/* * Author: sweat123 * Created Time: 2016/7/12 10:39:09 * File Name: main.cpp */ #include<set> #include<map> #include<queue> #include<stack> #include<cmath> #include<string> #include<vector> #include<cstdio> #include<time.h> #include<cstring> #include<iostream> #include<algorithm> #define INF 1<<30 #define MOD 1000000007 #define ll long long #define lson l,m,rt<<1 #define key_value ch[ch[root][1]][0] #define rson m+1,r,rt<<1|1 #define pi acos(-1.0) using namespace std; const int MAXN = 200010; int a[MAXN],n,ans[MAXN]; struct node{ int len; int val; node(){} node(int tlen,int tval):len(tlen),val(tval){} }; stack<node>s; int main(){ while(~scanf("%d",&n)){ for(int i = 1; i <= n; i++){ scanf("%d",&a[i]); } while(!s.empty())s.pop(); memset(ans,0,sizeof(ans)); for(int i = 1; i <= n; i++){ int len = 0;//表示当前弹出的这些元素能够贡献多少的长度 while(!s.empty()){ node tp = s.top(); if(tp.val < a[i])break; s.pop(); int ret = tp.len + len;//这个元素能够贡献的长度 （也就是说这个元素已经统计过了多少长度，在这里也是可以连续的） ans[ret] = max(ans[ret],tp.val); len += tp.len; } s.push(node(len+1,a[i])); } int len = 0; while(!s.empty()){ node tp = s.top(); s.pop(); int ret = len + tp.len; ans[ret] = max(ans[ret],tp.val); len += tp.len; } 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("\n"); } return 0; }
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原文地址：http://www.cnblogs.com/sweat123/p/5662894.html
