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],…,an,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]
问有多少个团体，团体要求能力最大和最小的差距不超过k，rmq,因为区间越长，最大值越大最小值越小，所以可以用二分找到左端点固定，右端点最右的区间，就可以了
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#define sf scanf
#define pf printf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define prf(x) printf("%d
",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
using namespace std;
const double pi=acos(-1.0);
const int N=1e5+10;
int a[N];
int dpmax[N][20],dpmin[N][20];
void first(int n)
{
mm(dpmax,0);
mm(dpmin,0);
rep(i,1,n+1)
{
dpmin[i][0]=dpmax[i][0]=a[i];
}
for(int j=1;(1<<j)<=n;j++)
{
for(int i=1;i+(1<<j)-1<=n;i++)
{
dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]);
dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]);
}
}
}
int fmax(int l,int r)
{
int x=0;
while(l-1+(1<<x)<=r) x++;
x--;
return max(dpmax[l][x],dpmax[r-(1<<x)+1][x]);
}
int fmin(int l,int r)
{
int x=0;
while(l-1+(1<<x)<=r) x++;
x--;
return min(dpmin[l][x],dpmin[r-(1<<x)+1][x]);
}
bool judge(int l,int mid,int k)
{
return fmax(l,mid)-fmin(l,mid)<k;
}
int main()
{
int re;scf(re);
while(re--)
{
int n,k;scff(n,k);
rep(i,1,n+1) scf(a[i]);
first(n);
ll ans=0;
rep(i,1,n)
{
int l=i,r=n,mid;
while(r-l>1)
{
mid=(l+r)/2;
if(judge(i,mid,k))
l=mid;
else
r=mid;
}
while(judge(i,l,k)&&l<=n)
l++;
l--;
ans+=l-i+1;
}
ans++;
pf("%lld
",ans);
}
}
还有一种用deque的方法 ，速度更快,慢慢体会吧...
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define prf(x) printf("%d
",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<deque>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
using namespace std;
const double pi=acos(-1.0);
const int N=1e5+10;
int a[N];
deque<int>v1;
deque<int>v2;
int main()
{
int re,n,k;
scf(re);
while(re--)
{
ll ans=0;
scff(n,k);
rep(i,1,n+1)
scf(a[i]);
v1.clear();
v2.clear();
int i,j;
for( i=1,j=1;i<=n;i++)
{
while(!v1.empty()&&v1.back()<a[i]) v1.pop_back();
v1.push_back(a[i]);
while(!v2.empty()&&v2.back()>a[i]) v2.pop_back();
v2.push_back(a[i]);
while(!v1.empty() &&!v2.empty() &&(v1.front()-v2.front()>=k))
{
ans+=i-j;
if(v1.front()==a[j]) v1.pop_front();
if(v2.front()==a[j]) v2.pop_front();
j++;
}
}
while(j<=n)
{
ans+=i-j;
j++;
}
cout<<ans<<endl;
}
return 0;
}