codeforces 631C (单调栈+思维)

C. Report

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are n commodities produced by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers. Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers in non-descending or non-ascending order and then passes the report to the manager i + 1, or directly to Blake (if this manager has number i = m).

Employees of the "Blake Technologies" are preparing the report right now. You know the initial sequence ai of length n and the description of each manager, that is value ri and his favourite order. You are asked to speed up the process and determine how the final report will look like.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of commodities in the report and the number of managers, respectively.

The second line contains n integers ai (|ai| ≤ 109) — the initial report before it gets to the first manager.

Then follow m lines with the descriptions of the operations managers are going to perform. The i-th of these lines contains two integers tiand ri (, 1 ≤ ri ≤ n), meaning that the i-th manager sorts the first ri numbers either in the non-descending (if ti = 1) or non-ascending (if ti = 2) order.

Output

Print n integers — the final report, which will be passed to Blake by manager number m.

Examples

input

3 11 2 32 2

output

2 1 3 

input

4 21 2 4 32 31 2

output

2 4 1 3 

Note

In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2 1 3. The report got to Blake in this form.

In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4 2 1 3. After the second manager the report changed to: 2 4 1 3. This report was handed over to Blake.

题意：

给出一个长为n的序列和m次操作，指令r中1表示排非减增序列，2表示排非增减序列，t表示从1到位置t的数按照指令r的要求排序。

思路：

对于排序来说如果t大于前面的t那么，前面的排序都是无用的。知道了这点就可以用一个单调栈来维护操作，使得最后需要进行的操作都是必须的且t按降序排列。

而对于后面的操作，必须要在前面已经排好的序列下进行，所以必须是从前向后进行操作。而t是从大到小排序的，所以当t减小时，就能确定一些数不在进行任何的排序了。所以每次只要确定下来不在排序的数即可。这样时间复杂度可以降到O(n)。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=200005;struct stru{    int r,t;}p[maxn],tmp;bool cmp(int a,int b){    return a>b;}int main(){    int n,m;    int a[maxn],b[maxn];    while(~scanf("%d%d",&n,&m))    {        int top=0,siz=0;        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        for(int i=1;i<=m;i++)        {            scanf("%d%d",&tmp.r,&tmp.t);            while(siz>0&&p[top].t<tmp.t)            {                top--;                siz--;            }            if(siz>0&&p[top].r==tmp.r)                continue;            p[++top]=tmp;            siz++;        }        if(p[1].r==1)            sort(a+1,a+1+p[1].t);        else            sort(a+1,a+1+p[1].t,cmp);        for(int i=1;i<=n;i++)            b[i]=a[i];        int r=p[1].t,l=1;        if(p[1].r==1)//l和r怎么对应b数组与一开始升序还是降序有关        {            for(int i=2;i<=siz;i++)            {                if(p[i-1].r==1)                {                    for(int j=p[i-1].t;j>p[i].t;j--)                    {                        b[j]=a[r--];                    }                }                else                {                    for(int j=p[i-1].t;j>p[i].t;j--)                    {                        b[j]=a[l++];                    }                }            }            if(p[siz].r==1)                for(int i=p[siz].t;i>=1;i--)                {                    b[i]=a[r--];                }            else                for(int i=p[siz].t;i>=1;i--)                {                    b[i]=a[l++];                }        }        else        {            for(int i=2;i<=siz;i++)            {                if(p[i-1].r==1)                {                    for(int j=p[i-1].t;j>p[i].t;j--)                    {                        b[j]=a[l++];                    }                }                else                {                    for(int j=p[i-1].t;j>p[i].t;j--)                    {                        b[j]=a[r--];                    }                }            }            if(p[siz].r==1)                for(int i=p[siz].t;i>=1;i--)                {                    b[i]=a[l++];                }            else                for(int i=p[siz].t;i>=1;i--)                {                    b[i]=a[r--];                }        }        for(int i=1;i<=n;i++)        {            if(i==1)                printf("%d",b[i]);            else                printf(" %d",b[i]);        }        printf("\n");    }    return 0;}