XKC's basketball team（单调栈+二分）

XKC , the captain of the basketball team , is directing a train of nnn team members. He makes all members stand in a row , and numbers them 1⋯n1 \cdots n1⋯n from left to right.

The ability of the iii-th person is wiw_iwi​ , and if there is a guy whose ability is not less than wi+mw_i+mwi​+m stands on his right , he will become angry. It means that the jjj-th person will make the iii-th person angry if j>ij>ij>i and wj≥wi+mw_j \ge w_i+mwj​≥wi​+m.

We define the anger of the iii-th person as the number of people between him and the person , who makes him angry and the distance from him is the longest in those people. If there is no one who makes him angry , his anger is −1-1−1 .

Please calculate the anger of every team member .

Input

The first line contains two integers nnn and m(2≤n≤5∗105,0≤m≤109)m(2\leq n\leq 5*10^5, 0\leq m \leq 10^9)m(2≤n≤5∗105,0≤m≤109) .

The following  line contain nnn integers w1..wn(0≤wi≤109)w_1..w_n(0\leq w_i \leq 10^9)w1​..wn​(0≤wi​≤109) .

Output

A row of nnn integers separated by spaces , representing the anger of every member .

样例输入 复制

6 1
3 4 5 6 2 10

样例输出 复制

4 3 2 1 0 -1

解题思路：单调栈+二分 从后往前维护一个单调递增的栈  然后二分查找即可！

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n,m;
 4 const int maxn=5e5+5;
 5 int arr[maxn];
 6 int sta[maxn];
 7 int top=0;
 8 map<int,int> ma;
 9 
10 int Binary(int left,int right,int value){
11     int ans=-1;
12     while(left<=right){
13         int mid=left+right>>1;
14         if(arr[sta[mid]]>=value) ans=mid,right=mid-1;
15         else left=mid+1;
16     }
17     return ans;
18 }
19 
20 int main(){
21     ios::sync_with_stdio(false);
22     cin>>n>>m;
23     for(int i=1;i<=n;i++) cin>>arr[i];
24     int top=1;
25     sta[top]=n;ma[n]=-1;
26     for(int i=n-1;i>=1;i--){
27         if(arr[i]>arr[sta[top]]-m) ma[i]=-1;
28         else{
29             int flag=Binary(1,top,arr[i]+m);   //找大于等于这个数的最左边的位置
30 //            cout << arr[sta[flag]] << endl;
31             ma[i]=sta[flag]-i-1;
32         }
33         if(arr[i]>arr[sta[top]]) sta[++top]=i;
34     }
35     for(int i=1;i<=n;i++){
36         printf("%d%c",ma[i],i==n?'\n':' ');
37     }
38     return 0;
39 }
40 
41 //10 1
42 //3 4 5 6 2 20 10 30 40 50

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