解题报告 之 CodeForces 6E Exposition

解题报告 之 CodeForces 6E Exposition

Description

There are several days left before the fiftieth birthday of a famous Berland's writer Berlbury. In this connection the local library decided to make an exposition of the works of this famous science-fiction writer. It was decided as well that it is necessary to include into the exposition only those books that were published during a particular time period. It is obvious that if the books differ much in size, the visitors will not like it. That was why the organizers came to the opinion, that the difference between the highest and the lowest books in the exposition should be not more than k millimeters.

The library has n volumes of books by Berlbury, arranged in chronological order of their appearance. The height of each book in millimeters is know, it is h_i. As Berlbury is highly respected in the city, the organizers want to include into the exposition as many books as possible, and to find out what periods of his creative work they will manage to cover. You are asked to help the organizers cope with this hard task.

Input

The first line of the input data contains two integer numbers separated by a space n (1 ≤ n ≤ 10⁵) and k (0 ≤ k ≤ 10⁶) — the amount of books by Berlbury in the library, and the maximum allowed height difference between the lowest and the highest books. The second line contains n integer numbers separated by a space. Each number h_i (1 ≤ h_i ≤ 10⁶) is the height of the i-th book in millimeters.

Output

In the first line of the output data print two numbers a and b (separate them by a space), where a is the maximum amount of books the organizers can include into the exposition, and b — the amount of the time periods, during which Berlbury published a books, and the height difference between the lowest and the highest among these books is not more than k milllimeters.

In each of the following b lines print two integer numbers separated by a space — indexes of the first and the last volumes from each of the required time periods of Berlbury's creative work.

Sample Input

Input

3 3
14 12 10

Output

2 2
1 2
2 3

Input

2 0
10 10

Output

2 1
1 2

Input

4 5
8 19 10 13

Output

2 1
3 4

题目大意：给一个n个元素的序列，从中挑出最长的子序列，要求子序列中元素差的最大值不超过k。问有几个最长子序列，子序列长度，以及这几个子序列的起始、终止位置。

分析：不会单调队列的朋友先看看我的另一篇博文，题目跟着道题其实差不多。（http://blog.youkuaiyun.com/maxichu/article/details/47612497）

我们用单调队列动态维护序列中的最大值和最小值。设立前后指针i、j，一旦发现前指针走到了最大值-最小值>k的地方，就试图更新一下最大值，然后 j 再往前走，以此类推。

上代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<deque>
#include<utility>
#include<vector>
using namespace std;

const int MAXN = 1e5 + 10;

int book[MAXN];
vector<pair<int, int> > p;
deque<int> high;
deque<int> low;

int main()
{
	int n, k;
	while(cin >> n >> k)
	{
		for(int i = 1; i <= n; i++)
		{
			scanf( "%d", &book[i] );
		}

		p.clear();
		high.clear();
		low.clear();

		int i, j; //a is pre pointer
		int ans = 0, cnt = 0;
		for(i = j = 1; i <= n; i++)
		{

			while(!high.empty()&&high.back() < book[i]) high.pop_back();
			high.push_back( book[i] );

			while(!low.empty()&&low.back() > book[i]) low.pop_back();
			low.push_back( book[i] );

			while(!high.empty() && !low.empty() && high.front() - low.front()>k)
			{
				if(p.empty() || i - j >ans)
				{
					p.clear();
					cnt = 1;
					ans = i - j;
					p.push_back( make_pair( j, i - 1 ) );
				}

				else if(!p.empty() && i - j == ans)
				{
					cnt++;
					p.push_back( make_pair( j, i - 1 ) );
				}
				if(high.front() == book[j]) high.pop_front();
				if(low.front() == book[j]) low.pop_front();
				j++;
			}
		}

		while(j <=n)
		{
			if(p.empty() || i - j >ans)
			{
				p.clear();
				cnt = 1;
				ans = i - j;
				p.push_back( make_pair( j, i - 1 ) );
			}

			else if(!p.empty() && i - j == ans)
			{
				cnt++;
				p.push_back( make_pair( j, i - 1 ) );
			}
			if(high.front() == book[j]) high.pop_front();
			if(low.front() == book[j]) low.pop_front();
			j++;

		}
		
		cout << ans << " " << cnt << endl;
		for(int i = 0; i < p.size(); i++)
		{
			cout << p[i].first << " " << p[i].second << endl;
		}
	}
	return 0;
}