Hdu-5289 Assignment (二分+RMQ || 单调队列)-优快云博客

本文介绍了一种计算数列中满足特定条件子序列数量的方法，提供了两种不同的算法实现方案：一是利用二分查找和预处理技巧；二是使用双端单调队列来高效地更新和维护子序列的最大和最小值。

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Problem Description

Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In a group, the difference of the ability of any two staff is less than k, and their numbers are continuous. Tom want to know the number of groups like this.

Input

In the first line a number T indicates the number of test cases. Then for each case the first line contain 2 numbers n, k (1<=n<=100000, 0<k<=10^9),indicate the company has n persons, k means the maximum difference between abilities of staff in a group is less than k. The second line contains n integers:a[1],a[2],…,a[n](0<=a[i]<=10^9),indicate the i-th staff’s ability.

Output

For each test，output the number of groups.

Sample Input

  
   2
4 2
3 1 2 4
10 5
0 3 4 5 2 1 6 7 8 9

Sample Output

  
   5
28

   
    
     Hint
    First Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]

Author

FZUACM

Source

2015 Multi-University Training Contest 1

题意：给你一个数列，问你它有多少个子序列满足最大值-最小值小于k.

解法1：枚举起始位置，二分最大有效长度。

#include <cstdio>
#include <iostream>
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) > (b) ? (b) : (a)
using namespace std;
int n,k,T,l[100005][30],r[100005][30],a[100005];
int Log(int x)
{
	int num = 0;
	while(x)
	{
		num++;
		x = x>>1;
	}
	return num-1;
}
int got(int x)
{
	return 1<<x;
}
int gotmax(int x,int y)
{
	int L = Log(y-x+1);
	return max(r[x][L],r[y-got(L)+1][L]);
}
int gotmin(int x,int y)
{
	int L = Log(y-x+1);
	return min(l[x][L],l[y-got(L)+1][L]);
}
bool check(int x,int y)
{
	if(gotmax(x,y)-gotmin(x,y) < k) return true;
	return false;
}
int main() 
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d %d",&n,&k);
		long long ans = 0;
		for(int i = 1;i <= n;i++) 
		{
			scanf("%d",&a[i]);
			l[i][0] = a[i];
			r[i][0] = a[i];
		}		
		for(int i = 1;got(i) <= n;i++)
		 for(int j = 1;j+got(i)-1 <= n;j++)
		  {
		  	l[j][i] = min(l[j][i-1],l[j+got(i-1)][i-1]);
		  	r[j][i] = max(r[j][i-1],r[j+got(i-1)][i-1]);
		  }
 		for(int i = 1;i <= n;i++)
		{
			int s = i,t = n;
			while(s != t)
			{
				int mid = (s+t)/2 + 1;
				if(check(i,mid)) s = mid;
				else t = mid-1;
			}
			ans += (s-i+1ll);
		}
		cout<<ans<<endl; 
	}
}

解法2：双端单调队列维护最大最小值。

#include <queue>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int n,k,a[100007],T;
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		long long ans = 0;
		deque <int> Max,Min;
		scanf("%d %d",&n,&k);
		for(int i = 1;i <= n;i++) scanf("%d",&a[i]);
		int j = 1;
		for(int i = 1;i <= n;i++)
		{
			while(!Max.empty() && Max.front() < i) Max.pop_front();
			while(!Min.empty() && Min.front() < i) Min.pop_front();
			while(j <=n && (Max.empty() || (abs(a[Max.front()]-a[j]) < k && abs(a[Min.front()]-a[j]) < k)))
			{
				while(!Max.empty() && a[Max.back()] <= a[j]) Max.pop_back();
				while(!Min.empty() && a[Min.back()] >= a[j]) Min.pop_back();
				Max.push_back(j);
				Min.push_back(j);
				j++;
			}	
			ans += j-i;
		}
		cout<<ans<<endl;
	}
}