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

最新推荐文章于 2024-12-09 01:52:22 发布

原创最新推荐文章于 2024-12-09 01:52:22 发布 · 398 阅读

0 ·

CC 4.0 BY-SA版权

codeforces 专栏收录该内容

48 篇文章

订阅专栏

本文介绍了一种高效的报告排序算法，该算法能够快速处理一系列经济指标，并通过特定操作序列优化报告呈现，确保最终报告准确且高效地传递给公司高层。

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, thei-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 ( $\text{[math]}$ , 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 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:

给你一个长度为n的初始的序列和q个操作((1 ≤ n,q ≤ 200000),一个操作包括ti和ri,ti为1表示从小到大,ti为2表示从大到小,ri表示对前面多少个进行排序。

思路:如果后面的操作长度比前面的长度要大,那么可以知道前面的操作便是无效的,那么如果前面的操作是有效的话,他比后面操作最大的序列长度要大,那么他比后面的最大的序列长度大的那一部分便可以直接确定了,因为有效的操作序列一定是单调递减的。

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
const int M=1e9+7;
const int maxn=200100;
int a[maxn];

bool cmp(int x,int y){
    return x>y;
}

struct node{
    int op;
    int id;
}line[maxn];
int num[maxn];
int C[maxn];

int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        int minv=0;
        for(int i=1;i<=m;i++){
            scanf("%d%d",&line[i].op,&line[i].id);
            minv=max(line[i].id,minv);
        }
        int maxv=0;
        for(int i=m;i>=1;i--){
            num[i]=maxv;
            maxv=max(maxv,line[i].id);
        }
        for(int i=1;i<=maxv;i++)
            C[i]=a[i];
        int st=1,ed=maxv;
        sort(C+1,C+maxv+1);
        for(int i=1;i<=m;i++){
            if(line[i].id<=num[i])
                continue;
            if(line[i].op==1){  //xiao dao da
                int j,k;
                for(j=0;line[i].id-j>num[i];j++){
                    a[line[i].id-j]=C[ed];
                    ed--;
                }
            }
            else{   //da dao xiao
                int j;
                for(j=0;line[i].id-j>num[i];j++){
                    a[line[i].id-j]=C[st];
                    st++;
                }
            }
        }
        for(int i=1;i<=n;i++){
            if(i==1)
                printf("%d",a[i]);
            else
                printf(" %d",a[i]);
        }
        printf("\n");
    }
    return 0;
}