codeforce 631C report

Report

limit 2s,256M

    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 ≤ 200000) — 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

4 2

1 2 4 3

2 3

1 2

output

2 4 1 3

題意

給定一個序列，進行m次操作，每一次都會把區間[1-r]内的數升序或者降序排列求m次操作之後的序列

題解（此處照搬我校卿神題解）

#include<stdio.h>
#include<iostream>
#include<algorithm>

using namespace std;
const int maxn=2e5;

struct Struct
{
	int time;
	int x;
	int op;
};

Struct opr[maxn+5];
int arr[maxn+5];
int ans[maxn+5];

bool cmp(const Struct &x,const Struct &y)
{
	return x.x<y.x;
}

bool op2(const int &a,const int &b)
{
	return a>b;
}

int main(void)
{
	#ifdef ex
	freopen ("in.txt","r",stdin);
	//freopen ("1.txt","w",stdout);
	#endif
	
	int n,m;
	int t1,t2;
	
	scanf("%d%d",&n,&m);
	
	for (int i=1;i<=n;++i)
	{
		scanf("%d",&arr[i]);
	}
	for (int i=1;i<=m;++i)
	{
		scanf("%d%d",&t1,&t2);
		opr[i].op=t1;
		opr[i].x=t2;
		opr[i].time=i;
	}
	
	sort(opr+1,opr+m+1,cmp);
	
	int cnt=m;
	int now_op;
	
	if (opr[cnt].op==1) 
	{
		sort(arr+1,arr+opr[cnt].x+1);
		now_op=1;
	}
	else 
	{
		sort(arr+1,arr+opr[cnt].x+1,op2);
		now_op=2;
	}
	
	for (int i=opr[m].x+1;i<=n;++i)
	{
		ans[i]=arr[i];
	}
	
	int s=1;
	int t=opr[m].x;
	int k=t;
	opr[0]=(Struct){m+1,0,3};
	for (int i=cnt;i>=0;--i)
	{
		//if (opr[i].time>opr[cnt].time && opr[i].op!=opr[cnt].op)
		if (opr[i].time>opr[cnt].time )
		{
			int len=opr[cnt].x-opr[i].x;
			if (now_op==opr[m].op)
			{
				for (int j=1;j<=len;++j)
				{
					ans[k]=arr[t];
					--k;
					--t;
				}
			}
			else
			{
				for (int j=1;j<=len;++j)
				{
					ans[k]=arr[s];
					++s;
					--k;
				}
			}
			cnt=i;
			now_op=opr[cnt].op;
		}
	}
	
	for (int i=1;i<=n;++i)
	{
		printf("%d ",ans[i]);
	}
	printf("\n");
}
           

有一處細節

if (opr[i].time>opr[cnt].time && opr[i].op!=opr[cnt].op)是錯誤的。雖然同樣的op不會影響目前序列的複制，但是如果不更新time就會導緻後面出錯

if (opr[i].time>opr[cnt].time )才是正确的