You‘re given a permutation aaa of length nnn (1≤n≤1051 \le n \le 10^51≤n≤105).
For each i∈[1,n]i \in [1,n]i∈[1,n], construct a sequence sis_isi? by the following rules:
- si[1]=is_i[1]=isi?[1]=i;
- The length of sis_isi? is nnn, and for each j∈[2,n]j \in [2, n]j∈[2,n], si[j]≤si[j−1]s_i[j] \le s_i[j-1]si?[j]≤si?[j−1];
- First, we must choose all the possible elements of sis_isi? from permutation aaa. If the index of si[j]s_i[j]si?[j] in permutation aaa is pos[j]pos[j]pos[j], for each j≥2j \ge 2j≥2, ∣pos[j]−pos[j−1]∣≤k|pos[j]-pos[j-1]|\le k∣pos[j]−pos[j−1]∣≤k (1≤k≤1051 \le k \le 10^51≤k≤105). And for each sis_isi?, every element of sis_isi? must occur in aaa at most once.
- After we choose all possible elements for sis_isi?, if the length of sis_isi? is smaller than nnn, the value of every undetermined element of sis_isi? is 000;
- For each sis_isi?, we must make its weight high enough.
Consider two sequences C=[c1,c2,...cn]C = [c_1, c_2, ... c_n]C=[c1?,c2?,...cn?] and D=[d1,d2,...,dn]D=[d_1, d_2, ..., d_n]D=[d1?,d2?,...,dn?], we say the weight of CCC is higher than that of DDD if and only if there exists an integer kkk such that 1≤k≤n1 \le k \le n1≤k≤n, ci=dic_i=d_ici?=di? for all 1≤i<k1 \le i < k1≤i<k, and ck>dkc_k > d_kck?>dk?.
If for each i∈[1,n]i \in [1,n]i∈[1,n], ci=dic_i=d_ici?=di?, the weight of CCC is equal to the weight of DDD.
For each i∈[1,n]i \in [1,n]i∈[1,n], print the number of non-zero elements of sis_isi? separated by a space.
It‘s guaranteed that there is only one possible answer.
Input
There are multiple test cases.
The first line contains one integer T(1≤T≤20)T(1 \le T \le 20)T(1≤T≤20), denoting the number of test cases.
Each test case contains two lines, the first line contains two integers nnn and kkk (1≤n,k≤1051 \le n,k \le 10^51≤n,k≤105), the second line contains nnn distinct integers a1,a2,...,ana_1, a_2, ..., a_na1?,a2?,...,an? (1≤ai≤n1 \le a_i \le n1≤ai?≤n) separated by a space, which is the permutation aaa.
Output
For each test case, print one line consists of nnn integers ∣s1∣,∣s2∣,...,∣sn∣|s_1|, |s_2|, ..., |s_n|∣s1?∣,∣s2?∣,...,∣sn?∣ separated by a space.
∣si∣|s_i|∣si?∣ is the number of non-zero elements of sequence sis_isi?.
There is no space at the end of the line.
样例输入 复制
2 3 1 3 2 1 7 2 3 1 4 6 2 5 7
样例输出 复制
1 2 3 1 1 2 3 2 3 3解题思路: 题目意思找[x-k,x+k]区间里小于其的最大值,然后在用那个数去更新 位置一个单调递减的数列！
1 #include <bits/stdc++.h>
2 using namespace std;
3 int tot=0,t,n,k;
4 const int maxn=1e5+5;
5 struct Node{
6 int l,r,sum;
7 }A[maxn*32];
8 int arr[maxn],root[maxn];
9 int ma[maxn],res[maxn];
10
11 void updata(int left,int right,int &now,int pre,int value){
12 A[++tot]=A[pre],now=tot,A[now].sum++;
13 if(left==right) return;
14 int mid=left+right>>1;
15 if(mid>=value) updata(left,mid,A[now].l,A[pre].l,value);
16 else updata(mid+1,right,A[now].r,A[pre].r,value);
17 }
18
19 int query(int left,int right,int x,int y,int value){ //返回的小于某个数的最大值
20 if(A[y].sum-A[x].sum==0) return -1; //当前区间的个数为0则减枝;
21 if(left==right){ //递归到叶子节点
22 return left<value?left:-1;
23 }
24 int mid=left+right>>1;
25 if(value<=mid+1||A[A[y].r].sum-A[A[x].r].sum==0){ //查找的数在左边或者右边没有shu
26 return query(left,mid,A[x].l,A[y].l,value);
27 }
28 int t=query(mid+1,right,A[x].r,A[y].r,value); //先查找右边
29 if(t==-1) query(left,mid,A[x].l,A[y].l,value);
30 else return t;
31
32 }
33
34 int main(){
35 ios::sync_with_stdio(false);
36 cin>>t;
37 while(t--){
38 memset(root,0,sizeof(root)),tot=0,memset(A,0,sizeof(A));
39 cin>>n>>k;
40 for(int i=1;i<=n;i++) cin>>arr[i],ma[arr[i]]=i;
41 for(int i=1;i<=n;i++) updata(1,n,root[i],root[i-1],arr[i]);
42 res[1]=1;
43 for(int i=2;i<=n;i++){
44 int L=max(1,ma[i]-k);
45 int R=min(n,ma[i]+k);
46 int flag=query(1,n,root[L-1],root[R],i);
47 if(flag==-1) res[i]=1;
48 else res[i]=res[flag]+1; //递推的公式
49 }
50 for(int i=1;i<=n;i++){
51 printf("%d%c",res[i],i==n?‘\n‘:‘ ‘);
52 }
53 }
54 return 0;
55 }
主席树
1 #include <bits/stdc++.h>
2 using namespace std;
3 int t,n,m;
4 const int maxn=1e5+5;
5 int sum[maxn<<2];
6 int ma[maxn],res[maxn],arr[maxn];
7 void updata(int rt) { sum[rt]=max(sum[rt<<1],sum[rt<<1|1]); } //区间存的是最大值
8 void build(int l,int r,int rt){
9 sum[rt]=0;
10 if(l==r) {return;}
11 int mid=l+r>>1;
12 build(l,mid,rt<<1);
13 build(mid+1,r,rt<<1|1);
14 updata(rt);
15 }
16 void add(int l,int r,int rt,int pos,int value){
17 if(l==r) {sum[rt]=value;return;}
18 int mid=l+r>>1;
19 if(mid>=pos) add(l,mid,rt<<1,pos,value);
20 else add(mid+1,r,rt<<1|1,pos,value);
21 updata(rt);
22 }
23 int query(int l,int r,int rt,int L,int R,int value){
24 if(L<=l&&R>=r) return sum[rt];
25 int mid=l+r>>1;
26 int ans=0;
27 if(mid>=L) ans=max(ans,query(l,mid,rt<<1,L,R,value));
28 if(R>mid) ans=max(ans,query(mid+1,r,rt<<1|1,L,R,value));
29 return ans;
30 }
31
32 int main(){
33 ios::sync_with_stdio(false);
34 cin>>t;
35 while(t--){
36 cin>>n>>m;
37 for(int i=1;i<=n;i++){
38 cin>>arr[i];
39 ma[arr[i]]=i;
40 }
41 build(1,n,1);
42 for(int i=1;i<=n;i++){
43 int L=max(1,ma[i]-m);
44 int R=min(n,ma[i]+m);
45 int x=query(1,n,1,L,R,i); //小于这个数的最大值
46 add(1,n,1,ma[i],i); //后添加必定保证是小于其的数
47 res[i]=res[x]+1;
48 }
49 for(int i=1;i<=n;i++)
50 printf("%d%c",res[i],i==n?‘\n‘:‘ ‘);
51 }
52 return 0;
53 }
线段树
原文地址：https://www.cnblogs.com/qq-1585047819/p/11517017.html