HDOJ 5289 Assignment 单调队列

维护一个递增的和递减的单调队列

Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 78    Accepted Submission(s): 40

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] 

 

Source

2015 Multi-University Training Contest 1

 

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/* ***********************************************
Author        :CKboss
Created Time  :2015年07月21日 星期二 12时36分35秒
File Name     :1002.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;
const int maxn=100100;

struct Node
{
    int val,pos;
};

int n,K;
int a[maxn];
// q1 dizheng q2 dijian
deque<Node> q1,q2;

LL ans;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        ans=0;
        while(!q1.empty()) q1.pop_back();
        while(!q2.empty()) q2.pop_back();

        scanf("%d%d",&n,&K);
        for(int i=0;i<n;i++) scanf("%d",a+i);

        int head=0;
        for(int i=0;i<n;i++)
        {
            Node node = (Node){a[i],i};

            /// push q1 tail dizheng
            while(!q1.empty())
            {
                Node b = q1.back();
                if(b.val<node.val) q1.pop_back();
                else break;
            }
            q1.push_back(node);

            /// push q2 tail dijian
            while(!q2.empty())
            {
                Node b = q2.back();
                if(b.val>node.val) q2.pop_back();
                else break;
            }
            q2.push_back(node);

            if(i==0) ans++;
            else
            {
                /// bijiao head
                while(true)
                {
                    Node big = q1.front();
                    Node small = q2.front();

                    if(big.val-small.val<K) break;
                    else
                    {
                        if(small.pos<big.pos)
                        {
                            head=small.pos+1; q2.pop_front();
                        }
                        else
                        {
                            head=big.pos+1; q1.pop_front();
                        }
                    }
                }
                ans+=i-head+1;
            }
        }

        cout<<ans<<endl;
    }

    return 0;
}