Greedy Sequence
限制
5000 ms
256 MB
You're given a permutation a of length n (1 ≤ n ≤ 105).
For each i ∈ [1, n], construct a sequence si by the following rules:
1. si[1] = i;
2. The length of si is n, and for each j ∈ [2, n], si[j] ≤ si[j − 1];
3. First, we must choose all the possible elements of si from permutation a. If the index
of si[j] in permutation a is pos[j], for each j ≥ 2, ∣pos[j] − pos[j − 1]∣ ≤ k (
1 ≤ k ≤ 105). And for each si, every element of si must occur in a at most once.
4. After we choose all possible elements for si, if the length of si is smaller than n, the
value of every undetermined element of si is 0;
5. For each si, we must make its weight high enough.
Consider two sequences C = [c1, c2, ...cn] and D = [d1, d2, ..., dn], we say the weight
of C is higher than that of D if and only if there exists an integer k such that 1 ≤ k ≤ n,
ci = di for all 1 ≤ i < k, and ck > dk .
If for each i ∈ [1, n], ci = di, the weight of C is equal to the weight of D.
For each i ∈ [1, n], print the number of non-zero elements of si separated by a space.It's guaranteed that there is only one possible answer.
Input
There are multiple test cases.
The fifirst line contains one integer T(1 ≤ T ≤ 20), denoting the number of test cases.
Each test case contains two lines, the fifirst line contains two integers n and k (
1 ≤ n, k ≤ 105), the second line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ n)
separated by a space, which is the permutation a.
Output
For each test case, print one line consists of n integers ∣s1∣, ∣s2∣, ..., ∣sn∣ separated by a
space.
∣si∣ is the number of non-zero elements of sequence si.
There is no space at the end of the line.
Sample Input 1
2
3 1
3 2 1
7 2
3 1 4 6 2 5 7
Sample Output 1
1 2 3
1 1 2 3 2 3 3
前面的ans对后面的ans有贡献，用 num [ i ] 存 len ( S i )，然后找到满足条件的 a[ j ] （即 abs（j - i）<=k ），len ( S i ) = 1 + num [ a[ j ] ]。
题目给的5000ms，1014ms就过了，嗨森。。（不过应该还可以更快，比赛的时候sb了，加了个memset。。。）
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn=1e6+7; 4 struct NODE 5 { 6 int data; 7 int id; 8 }node[maxn]; 9 bool cmp(NODE a,NODE b) 10 { 11 return a.data<b.data; 12 } 13 int num[maxn]; 14 int main() 15 { 16 int t; 17 scanf("%d",&t); 18 while(t--) 19 { 20 int n,k; 21 scanf("%d%d",&n,&k); 22 for(int i=1;i<=n;++i) 23 { 24 scanf("%d",&node[i].data); 25 node[i].id=i; 26 } 27 sort(node+1,node+n+1,cmp); 28 //memset(num,0,sizeof(num)); 29 for(int i=1;i<=n;++i) 30 { 31 int len=1; 32 int pos=node[i].id; 33 for(int j=i-1;j>=1;--j) 34 { 35 if(abs(pos-node[j].id)<=k) 36 { 37 len+=num[j]; 38 break; 39 } 40 } 41 num[i]=len; 42 printf("%d",len); 43 if(i<n)printf(" "); 44 else printf(" "); 45 } 46 } 47 return 0; 48 }