Codeforces548D Mike and Feet 数据结构+单调栈+优先队列

D - Mike and Feet

Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

SubmitStatus

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 from1 to n from left to right.i-th bear is exactly a_i 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 that1 ≤ x ≤ n the maximum strength among all groups of sizex.

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, ..., a_n (1 ≤ a_i ≤ 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

Input

101 2 3 4 5 4 3 2 1 6

Output

6 4 4 3 3 2 2 1 1 1

解题思路：

仔细的想了好久，发现这题就是一个单调栈问题，正好前段时间正好做过。

1，对于每个数要找他的影响范围，那么就是这个数左边第一个比他小的数，和右边第一个比他小的数字之间。

2，那么问题来了怎么找每一个数的这个区间呢，单调栈可以解决这个问题，参考我的博客

http://blog.csdn.net/fans_ac/article/details/51965804

还算详细的介绍了单调栈的问题，那么这个区间就很好的获取了。

3，获取区间了，怎么求得对于每一个1~k的区间的最大值呢。优先队列可以很好的解决这个问题。

4，优先队列每次都抛出一个最大值，这个最大值的区间就是这个区间的最大值，后面来的都没有大。

5，弹完队列里面所有的数据，就可以找到所有的结果了。

#include<cstdio>#include<cstring>#include<iostream>#include<stack>#include<queue>using namespace std;const int maxn = 200005 ;struct node{    int val;    int pos ;};struct node2{    int len ;    int val ;    bool operator < (const node2 & p)const{        return val<p.val ;    }} ;stack<node> qq ;int arry[maxn] ;int L[maxn] ;int R[maxn] ;priority_queue<node2> pp ;int main(){    int n ;    while(~scanf("%d",&n)){        for(int i=1;i<=n;i++){            scanf("%d",&arry[i]);        }        while(!qq.empty())qq.pop();        node def;        def.pos = 0 ;        def.val = -0x3f3f3f3f ;        //printf("eys");        qq.push(def);        for(int i = 1;i<=n;i++){            //printf("i = %d\n",i);            while(qq.top().val>=arry[i]){                qq.pop();            }            //printf("s\n");            node t ;            t.val = arry[i] ;            t.pos = i ;            L[i] = qq.top().pos+1;            qq.push(t);        }        //printf("eys");        while(!qq.empty())qq.pop();        def.pos = n+1 ;        def.val = -0x3f3f3f3f ;        qq.push(def);        for(int i = n;i>=1;i--){            while(qq.top().val>=arry[i]){                qq.pop();            }            node t ;            t.val = arry[i] ;            t.pos = i ;            R[i] = qq.top().pos-1;            qq.push(t);        }//        for(int i =1;i<=n;i++){//            printf("%d %d\n",L[i],R[i]);//        }        while(!pp.empty())pp.pop();        for(int i = 1;i<=n;i++){            node2 newnode ;            newnode.len = R[i] - L[i] + 1 ;            newnode.val = arry[i] ;            pp.push(newnode) ;        }        int ans[maxn];        memset(ans,0,sizeof(ans));        while(!pp.empty()){            for(int i=pp.top().len;i>=1;i--){                if(ans[i]!=0)break ;                else ans[i]=pp.top().val ;            }            pp.pop();        }        printf("%d",ans[1]);        for(int i = 2;i<=n;i++){            printf(" %d",ans[i]);        }printf("\n");    }    return 0;}