Codeforces Round #344 (Div. 2) C. Report

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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 manageri + 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.




题目大意:给一个序列,进行m次操作,每次操作对[1,r]进行升序or降序,输出最终序列。


思路:不难发现,对于每个区间 [ri+1+1,ri]    只有第i次操作对此区间有作用,并且当

ri>rj,i>j  时,第j次操作其实是没有用的,因此,用两个数组op和t,

op[y]=x,表示对区间[1,y]的操作是x;

t[y]=i,表示对区间[1,y]的操作是第i次;

然后假设所有操作r最大为r_max,那么就从r_max开始倒序维护区间。



/* ***********************************************┆  ┏┓   ┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃       ┃ ┆┆┃   ━   ┃ ┆┆┃ ┳┛ ┗┳ ┃ ┆┆┃       ┃ ┆┆┃   ┻   ┃ ┆┆┗━┓ 马 ┏━┛ ┆┆  ┃ 勒 ┃  ┆      ┆  ┃ 戈 ┗━━━┓ ┆┆  ┃ 壁     ┣┓┆┆  ┃ 的草泥马  ┏┛┆┆  ┗┓┓┏━┳┓┏┛ ┆┆   ┃┫┫ ┃┫┫ ┆┆   ┗┻┛ ┗┻┛ ┆************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>using namespace std;#define rep(i,a,b) for (int i=(a),_ed=(b);i<_ed;i++)#define per(i,a,b) for (int i=(b)-1,_ed=(a);i>=_ed;i--)#define inf 0x3f3f3f3f#define mod 1000000007#define ll long long#define ull unsigned long long#define N 200005int t[N],op[N],ans[N];int a[N];int n,m;int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int x,y;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    for(int i=1;i<=m;i++)    {        scanf("%d%d",&x,&y);        op[y]=x;        t[y]=i;    }    int len=n;    for(;len;len--)    {        if(op[len])break;        ans[len]=a[len];    }    sort(a+1,a+1+len);    int l=1,r=len;    int nowtime=0,nowop=0;    for(int i=len;i>=1;i--)    {        if(t[i]>nowtime)        {            nowtime=t[i];            nowop=op[i];        }        if(nowop==1)ans[i]=a[r--];        else ans[i]=a[l++];    }    for(int i=1;i<=n;i++)        printf("%d ",ans[i]);    printf("\n");    return 0;}

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