Hdu-5289 Assignment (二分+RMQ || 單調隊列)

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

題意：給你一個數列，問你它有多少個子序列滿足最大值-最小值小于k.

解法1：枚舉起始位置，二分最大有效長度。

#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：雙端單調隊列維護最大最小值。

#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;
	}
}