HDU 5089 Assignment（rmq+二分 或 单调队列）

Assignment

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

Total Submission(s): 557    Accepted Submission(s): 280

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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rmq+二分

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

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;
const int mod=1e9+7;

using namespace std;

#define N 100007

int a[N],n;
int dpmin[N][25],dpmax[N][25];
int k;

inline bool judge(int le,int ri)
{
    int kk=log2((ri-le+1)*1.0);
    int mi=min(dpmin[le][kk],dpmin[ri-(1<<kk)+1][kk]);
    int ma=max(dpmax[le][kk],dpmax[ri-(1<<kk)+1][kk]);
    return ma-mi<k;
}

int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
            dpmin[i][0]=dpmax[i][0]=a[i];

        for(j=1;(1<<j)<=n;j++)
            for(i=1;i+(1<<j)-1<=n;i++)
            {
                int p=1<<(j-1);
                dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+p][j-1]);
                dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+p][j-1]);
            }

         __int64 ans=0;
         int le,ri,p;
         for(i=1;i<=n;i++)
         {
             le=i;
             ri=n;
             while(le<=ri)
             {
                 int mid=(le+ri)>>1;
                 if(judge(i,mid))
                 {
                     p=mid;
                     le=mid+1;
                 }
                 else
                    ri=mid-1;
             }
             ans+=p-i+1;
         }
         printf("%I64d\n",ans);
    }
    return 0;
}

单调队列

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

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8

using namespace std;
const int mod=1e9+7;

#define INF 0x3f3f3f3f

const int N=100005;

int mique[N],maque[N],mihead,mahead,mitail,matail;
int n,a[N],k;

int pre,now;
__int64 ans;

inline void miinque(int i)
{
    while(mihead<mitail&&a[i]<a[mique[mitail-1]]) mitail--;
    mique[mitail++]=i;
}

inline void mainque(int i)
{
    while(mahead<matail&&a[i]>a[maque[matail-1]]) matail--;
      maque[matail++]=i;
}

void outque(int pos)
{
    if(a[maque[mahead]]-a[mique[mihead]]>=k)
    {
        int nowlen=pos-now-1;
        int prelen=pre-now;
        ans+=(__int64)(nowlen+1)*nowlen/2;
        if(prelen>=1)
            ans-=(__int64)(prelen+1)*prelen/2;
        pre=pos-1;
    }
    while(a[maque[mahead]]-a[mique[mihead]]>=k)
        if(mique[mihead]<maque[mahead])
        {
            now=mique[mihead]+1;
            mihead++;
        }
        else
        {
            now=maque[mahead]+1;
            mahead++;
        }
}

int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        mihead=mahead=mitail=matail=0;
        pre=now=1;
        ans=0;
        for(i=1;i<=n;i++)
        {
            miinque(i);
            mainque(i);
            outque(i);
        }
        if(pre<n)
        {
           int nowlen=n-now;
           int prelen=pre-now;
           ans+=(__int64)(nowlen+1)*(nowlen)/2;
           if(prelen>=1)
            ans-=(__int64)(prelen+1)*(prelen)/2;
           pre=n-1;
        }

        printf("%I64d\n",ans+n);
    }
    return 0;
}