E-XKC's basketball team
XKC , the captain of the basketball team , is directing a train of nn team members. He makes all members stand in a row , and numbers them 1 \cdots n1⋯n from left to right.
The ability of the ii-th person is w_iwi , and if there is a guy whose ability is not less than w_i+mwi+mstands on his right , he will become angry. It means that the jj-th person will make the ii-th person angry if j>ij>i and w_j \ge w_i+mwj≥wi+m.
We define the anger of the ii-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 .
Please calculate the anger of every team member .
Input
The first line contains two integers nn and 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 nn integers w_1..w_n(0\leq w_i \leq 10^9)w1..wn(0≤wi≤109) .
Output
A row of nn integers separated by spaces , representing the anger of every member .
样例输入
6 1
3 4 5 6 2 10
样例输出
4 3 2 1 0 -1
思路
标程用的是单调队列+二分，我用的是set+map，想法差不多。从后往前遍历arr[],严格单调递增插入arr[i],并用map将arr[i]的下标存起来,
ans[i]=lower_bound(arr[i]+m)==set.end()?-1:lower_bound(arr[i]+m)
1 #include <bits/stdc++.h> 2 #pragma GCC optimize(3) 3 using namespace std; 4 typedef long long ll; 5 const int maxn=1e6+7; 6 set<int> ste; 7 typedef set<int>::iterator sii; 8 map<int,int> MAP; 9 int ans[maxn]; 10 int arr[maxn]; 11 int main() 12 { 13 int n,m; 14 scanf("%d%d",&n,&m); 15 for(int i=1;i<=n;++i) 16 { 17 scanf("%d",&arr[i]); 18 } 19 for(int i=n;i>0;--i) 20 { 21 sii it=ste.lower_bound(arr[i]+m); 22 if(it!=ste.end()) 23 ans[i]=MAP[*it]-i-1; 24 else ans[i]=-1; 25 if(ste.empty()) 26 { 27 ste.insert(arr[i]); 28 MAP[arr[i]]=i; 29 } 30 else 31 { 32 it=--ste.end(); 33 if(arr[i]>(*it)) 34 { 35 ste.insert(arr[i]); 36 MAP[arr[i]]=i; 37 } 38 } 39 } 40 for(int i=1;i<=n;++i) 41 { 42 printf("%d",ans[i]); 43 if(i<n)printf(" "); 44 else printf("\n"); 45 } 46 return 0; 47 } 48 /* 49 5 2 50 5 4 3 2 1 51 */