hdu5289 Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2555    Accepted Submission(s): 1200

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]

这题有多种做法，但思路差不多，我的做法是依次枚举左端点或者右端点，然后二分查找所能达到的最右边，然后累加起来。

方法一：用树状数组或者rmq枚举左端点，然后二分查找满足条件的最右端点。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100060
#define ll long long
#define inf 2000000000
int a[maxn],b1[maxn],b2[maxn];/*b1最小值，b2最大值*/
int lowbit(int x){
    return x&(-x);
}
void update1(int pos,int num)
{
    while(pos<=maxn){
        b1[pos]=min(b1[pos],num);pos+=lowbit(pos);
    }
}

int question1(int pos)
{
    int num=inf;
    while(pos>0){
        num=min(b1[pos],num);pos-=lowbit(pos);
    }
    return num;
}

void update2(int pos,int num)
{
    while(pos<=maxn){
        b2[pos]=max(b2[pos],num);pos+=lowbit(pos);
    }
}

int question2(int pos)
{
    int num=-1;
    while(pos>0){
        num=max(b2[pos],num);pos-=lowbit(pos);
    }
    return num;
}

int main()
{
    int n,m,i,j,T,k,l,mid,r;
    ll ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        for(i=1;i<=n;i++){
            b1[i]=inf;b2[i]=-1;
            scanf("%d",&a[i]);
        }
        ans=0;
        for(i=n;i>=1;i--){
            update1(i,a[i]);
            update2(i,a[i]);
            l=i;r=n;
            while(l<=r){
                mid=(l+r)/2;
                if(question2(mid)-question1(mid)>=k){
                    r=mid-1;
                }
                else l=mid+1;
            }
            //printf("%d %d %d
",i,l,r);
            if(r>=i)
            ans+=(ll)(r-i+1);
            /*printf("%lld
",ans);*/
        }
        printf("%lld
",ans);
    }
    return 0;
}

代码二：维护两个单调队列，分别维护最大值和最小值，依次枚举右端点，然后找到符合条件的左端点。（这个方法速度是最快的，只需243ms，惊！）

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100060
#define ll long long
int q1[1111111][2],q2[1111111][2];
int a[maxn];
int main()
{
    int n,m,i,j,k,front1,front2,rear1,rear2,l,r,T;
    ll ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&k);
        ans=0;

        front1=front2=1;rear1=rear2=0;l=1;
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            while(front1<=rear1 && q1[rear1][0]>=a[i]){
                rear1--;
            }
            rear1++;q1[rear1][0]=a[i];q1[rear1][1]=i;

            while(front2<=rear2 && q2[rear2][0]<=a[i]){
                rear2--;
            }
            rear2++;q2[rear2][0]=a[i];q2[rear2][1]=i;
            while(front1<=rear1 && front2<=rear2 && q2[front2][0]-q1[front1][0]>=k){
                l=min(q2[front2][1],q1[front1][1])+1;
                if(q2[front2][1]<q1[front1][1]){
                    front2++;
                }
                else if(q2[front2][1]>q1[front1][1]){
                    front1++;
                }
                else{
                    front1++;front2++;
                }
            }
            ans+=i-l+1;
            /*printf("%d %d %lld
",i,l,ans);*/
        }
        printf("%lld
",ans);

    }
    return 0;
}