HDU 5289 Assignment（RMQ+二分）

Assignment

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

Total Submission(s): 3338    Accepted Submission(s): 1549


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
(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
HintFirst 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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代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;
int maxn[100000][30];
int minn[100000][30];
int a[100000];
int n,k;
void RMQ_init()
{
for(int j=1;(1<<j)<=n;j++)
{
for(int i=1;i+(1<<j)-1<=n;i++)
{
maxn[i][j]=max(maxn[i][j-1],maxn[i+(1<<(j-1))][j-1]);
minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);
}
}
}
int query(int l,int r)
{
int k=log2(r-l+1);
return max(maxn[l][k],maxn[r-(1<<k)+1][k])-min(minn[l][k],minn[r-(1<<k)+1][k]);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
maxn[i][0]=minn[i][0]=a[i];
}
RMQ_init();
long long ans=0;
int l,r;
for(int i=1;i<=n;i++)
{
l=i,r=n;
while(l<=r)
{
int mid=(l+r)/2;
int cha=query(i,mid);
if(cha<k)l=mid+1;
else r=mid-1;
}
ans=ans+l-i;
}
printf("%lld\n",ans);
}
}