HDU 5289

Assignment

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

Total Submission(s): 3478    Accepted Submission(s): 1605
[/b]

[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
For each test，output the number of groups.
 

[align=left]Sample Input[/align]

2
4 2
3 1 2 4
10 5
0 3 4 5 2 1 6 7 8 9

 

[align=left]Sample Output[/align]

5
28
HintFirst Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]

 

[align=left]Author[/align]
FZUACM
 

[align=left]Source[/align]
2015 Multi-University Training Contest 1

 

[align=left]Recommend[/align]
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算出有多少个区间。满足最大值-最小值 < k

解法：

枚举每个位置x是区间的最左边位置。二分查找，最右的位置r，满足[x,r]的最大值-最小值恰好 < k

r-x+1就是解

用树状数组更新最大最小值，二分位置。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include <cmath>
using namespace std;
#define maxn 100007
int tree1[maxn];
int tree2[maxn];
int lowbit(int i)
{
return i&(-i);
}
void add1(int i,int ans)
{
while(i<maxn)
{
tree1[i]=max(tree1[i],ans);
i+=lowbit(i);
}
}
void add2(int i,int ans)
{
while(i<maxn)
{
tree2[i]=min(tree2[i],ans);
i+=lowbit(i);
}
}
int add3(int i)
{
int res=1000000007;
while(i>0)
{
res=min(tree2[i],res);
i-=lowbit(i);
}
return res;
}
int add4(int i)
{
int res=0;
while(i>0)
{
res=max(tree1[i],res);
i-=lowbit(i);
}
return res;
}
int main()
{
int t,k,n;
cin>>t;
int a[maxn];
while(t--)
{
cin>>n>>k;
memset(tree1,0,sizeof(tree1));
memset(tree2,0x7f7f7f,sizeof(tree2));
for(int j=1; j<=n; j++)
cin>>a[j];
long long ans=0;
for(int j=n; j>=1; j--)
{
add1(j,a[j]);
add2(j,a[j]);
int low=j;
int high=n;
while(low<=high)
{
int mid=(low+high)/2;
int max1=add4(mid);
int min1=add3(mid);
if(max1-min1<k)
low=mid+1;
else
high=mid-1;
}
ans+=low-j;
}
cout<<ans<<endl;
}
return 0;
}