2015暑假多校联合---Assignment（优先队列）

原题链接

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

 

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题意：给了n个数和一个k，求有多少个区间满足区间中的极差小于看k；

 

思路：用两个优先队列q1、q2分别从小到大排列、从大到小排列（存的是下标），从i：0~n-1循环中，两个优先队列都push（i），这时可以得到当前区间的极差，即a[q2.top()]-a[q1.top()], 若小于k，则sum加上区间长度，区间开始位置注意用len1标记，结束位置即len2=i，那么sum+=len2-len1+1; 如果极差大于等于k，那么就要将区间长度缩短，删除一些优先队列中的数，首先判断：若q1.top()>q2.top()  则表示加入的数太小使该区间不符合要求，故删去区间前面太大的数，即删去q2中的数，先删去最大的数，那么最大的数之间的数也要删去，即t=q2.top() ;  

                                                  len1=t+1;

                                                  while（q2.top()<=t） q2.pop();

如果区间极差任然大于等于k，则继续重复上面操作，继续删除，直到区间符合，sum+=len2-len1+1; 若q1.top()<q2.top()  则和上面类似，删除q1中的太小的数；

 

代码如下：

#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstdio>
using namespace std;
int a[];

struct cmp1
{
    bool operator()(const int a1,const int a2)
    {
        return a[a1]>a[a2];
    }
};
struct cmp2
{
    bool operator()(const int a1,const int a2)
    {
        return a[a1]<a[a2];
    }
};
priority_queue<int,vector<int>,cmp1>q1;
priority_queue<int,vector<int>,cmp2>q2;

int main()
{
    int T;
    int n,k;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&k);
        for(int i=; i<n; i++)
            scanf("%d",&a[i]);
        long long sum=;
        int len1=,len2=;
        for(int i=; i<n; i++)
        {
            q1.push(i);
            q2.push(i);
            len2=i;
            endw:
            int maxn=a[q2.top()];
            int minn=a[q1.top()];
            if(maxn-minn<k) sum+=len2-len1+;
            else
            {
                if(q2.top()>q1.top())
                {
                    len1=t+;
                    int t=q1.top();
                    while(q1.top()<=t)
                      q1.pop();
                   goto endw;
                }
                else
                {
                    len1=t+;
                    int t=q2.top();
                    while(q2.top()<=t)
                        q2.pop();
                    goto endw;
                }
            }
        }
        while(!q1.empty())
            q1.pop();
        while(!q2.empty())
            q2.pop();
        printf("%lld\n",sum);
    }
    return ;
}