nefu1162

Single boy

Problem:1162

Time Limit:5000ms

Memory Limit:65535K

Description

Today is Christmas day. There are n single boys standing in a line. They are numbered form 1 to n from left to right. The i-th single boy has ai single strength. They are singing single boy Christmas song! Single boy,single boy, single all the way, having party together and turning into gay! Hey! A group of single boys is non-empty contiguous segment of the line. The size of a group is the number of single boys in that group. The single strength of a group is the minimum single strength of the single boy in that group.Now we want to know for each x such that 1<=x<=n the maximum single strength among all groups of size x.

Input

The first line of input contains T(<=30), the test cases.For each test case, the first line is a integer n(1 <= n <= 2*10^5), the number of single boys. The second line contains n integers [1,10^9] separated by space, the single strength of each single boy.

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

1101 2 9 3 8 2 2 10 19 6

Sample Output

19 10 6 2 2 2 2 2 2 1

Hint

Source

/**    题意：n个数,(1<=k<=n),在n个数中连续的k个数为一个区间长度，一共有n-k+1个区间，    每个区间选出区间中的最小值min[j](1<=j<=n-k+1)，然后在n-k+1个区间中选出最大的    min[j],得到max[i](1<=i<=k),然后按k=1~n顺序输出max[i].    由于数据太大，暴力模拟寻找太慢不能实现，所以我们想到的是求 以num[i]为最小值    的区间长度，然后选择区间长度相等Max(num[i]),开始我用了暴力寻找区间，由于数据    太大，TLE了，后来学习别人方法，用栈求得区间长度，我们把以num[i]为最小值的区间    的左右边界求出来，右边界-左边界就是区间长度；    用栈求区间左右边界：先求左边界后求右边界，开始将下标0入栈，我们的数组是从1    开始的，然后比较以栈顶元素为下标的值是否大于等于当前值，是的话出栈，直到比当前    值小，当前值的左边界就等于当前下标-栈顶元素+1，最后每次都要把当前下标入栈，    求右边界完全一样。*/#include <iostream>#include <stdio.h>#include <string.h>#include <stack>using namespace std;const int maxn=3000000;struct ac{    int num;    int l,r;}d[maxn];int ans[maxn];stack<int>sta;int main(){    int t,n,Top,dn;    scanf("%d",&t);    while(t--)    {        int flag=0;        scanf("%d",&n);        d[0].num=0;///以0 ，n+1为下标的值为0.        d[n+1].num=0;        sta.push(0);///压入0下标        for(int i=1; i<=n; i++)        {            scanf("%d",&d[i].num);            while(!sta.empty())///求左边界            {                Top=sta.top();                if(d[Top].num>=d[i].num)///比较当前值与以栈顶元素为下标的值                    sta.pop();                else                    break;            }            Top=sta.top();            d[i].l=Top+1;///左边界            sta.push(i);///每次压入当前下标        }        while(!sta.empty())        sta.pop();        sta.push(n+1);///求右边界，压入下标n+1        for(int i=n; i>=1; i--)        {            while(!sta.empty())            {                Top=sta.top();                if(d[Top].num>=d[i].num)                    sta.pop();                else                    break;            }            Top=sta.top();            d[i].r=Top-1;            sta.push(i);        }        while(!sta.empty())            sta.pop();        memset(ans,0,sizeof(ans));        for(int i=1; i<=n; i++)///求ans，每次都更新 以区间长度为dn的最大值        {              dn=d[i].r-d[i].l+1;              ans[dn]=max(ans[dn],d[i].num);        }        for(int i=n-1;i>=1;i--)///有些区间长度是找不到，那么用比他区间长度大的更新其值        {            if(ans[i]<ans[i+1])                ans[i]=ans[i+1];        }                printf("%d",ans[1]);        for(int i=2; i<=n; i++)            printf(" %d",ans[i]);        printf("\n");    }    return 0;}/**  样例值          1  2  9  3  8  2  2  10  19  6以该值为最小值的区间长度   10  9  1  3  1  9  9   2   1  3    区间长度为 10有1 ,max[10]=1;    区间长度为 9 有2,2,2,max[9]=2;    区间长度为 4-8 都没有，即为0,那么以区间为长度为9的更新max[4-8]=2;    区间长度为 3有 3,6, max[3]=6;    区间长度为 2有 10,  max[2]=10;      区间长度为 1有 9,8,19,max[1]=19;    为什么以区间更大的修改，以为区间越大，数值越小，区间小的的最大值不可能大于    区间大的最大值    */