【CodeForces

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 ti and 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.

Example
Input
3 1
1 2 3
2 2
Output
2 1 3
Input
4 2
1 2 4 3
2 3
1 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，给定一个序列，有m个人，每个人可以选择一种方式t将序列的前r个元素按照一定方式排序。

找规律，我们可以发现 按照时间顺序，只有当r不断减少的时候才是真正修改序列了，所以我们按照时间顺序维护一个非递减的的处理过程。
然后就是针对不断减少的r，最后输出的序列有什么特点，
可以发现，对于减少的两个r来说，差的那段区间数字的填充，是和大的r的操作有关，找到规律之后就不断填充 差的那段区间就行了。

#include<bits/stdc++.h>using namespace std;typedef pair<int,int> pii;#define fi first #define se second #define  LL long long#define fread() freopen("in.txt","r",stdin)#define fwrite() freopen("out.txt","w",stdout)#define CLOSE() ios_base::sync_with_stdio(false)const int MAXN = 200000+10;const int MAXM = 1e6;const int mod = 1e9+7;const int inf = 0x3f3f3f3f;int arr[MAXN],brr[MAXN];pii op[MAXN];int top=0; //  单调栈，维护一个非递减的操作序列int main(){    CLOSE();//  fread();//  fwrite();    int n,m;scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++) {        scanf("%d",&arr[i]);        brr[i]=arr[i];    }    while(m--){        int a,b;scanf("%d%d",&a,&b);        while(top&&b>op[top-1].se) top--;        op[top].fi=a;op[top].se=b;top++;    }    op[top].fi=1;op[top].se=0; top++;    sort(brr+1,brr+1+op[0].se);    int l=1,r=op[0].se;    for(int i=0;i<top-1;i++){        for(int j=op[i].se;j>=op[i+1].se+1;j--){            if(op[i].fi==1)  arr[j]=brr[r--];            else  arr[j]=brr[l++];              }    }     for(int i=1;i<=n;i++){        printf("%d",arr[i]);        if(i!=n) putchar(' ');    }    return 0;}