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(2leq nleq 5*10^5, 0leq 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(0leq 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！

#include <bits/stdc++.h>
using namespace std;
int n,m;
const int maxn=5e5+5;
int arr[maxn];
int sta[maxn];
int top=0;
map<int,int> ma;

int Binary(int left,int right,int value){
    int ans=-1;
    while(left<=right){
        int mid=left+right>>1;
        if(arr[sta[mid]]>=value) ans=mid,right=mid-1;
        else left=mid+1;
    }
    return ans;
}

int main(){
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=n;i++) cin>>arr[i];
    int top=1;
    sta[top]=n;ma[n]=-1;
    for(int i=n-1;i>=1;i--){
        if(arr[i]>arr[sta[top]]-m) ma[i]=-1;
        else{
            int flag=Binary(1,top,arr[i]+m);   //找大于等于这个数的最左边的位置
//            cout << arr[sta[flag]] << endl;
            ma[i]=sta[flag]-i-1;
        }
        if(arr[i]>arr[sta[top]]) sta[++top]=i;
    }
    for(int i=1;i<=n;i++){
        printf("%d%c",ma[i],i==n?'
':' ');
    }
    return 0;
}

//10 1
//3 4 5 6 2 20 10 30 40 50