A sequence S = {s1, s2, ..., sn} is called heapable if there exists a binary tree T with n nodes such that every node is labelled with exactly one element from the sequence S, and for every non-root node si and its parent sj, sj ≤ si and j < i hold. Each element in sequence S can be used to label a node in tree T only once.
Chiaki has a sequence a1, a2, ..., an, she would like to decompose it into a minimum number of heapable subsequences.
Note that a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contain an integer n (1 ≤ n ≤ 105) — the length of the sequence.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n).
It is guaranteed that the sum of all n does not exceed 2 × 106.
Output
For each test case, output an integer m denoting the minimum number of heapable subsequences in the first line. For the next m lines, first output an integer Ci, indicating the length of the subsequence. Then output Ci integers Pi1, Pi2, ..., PiCi in increasing order on the same line, where Pij means the index of the j-th element of the i-th subsequence in the original sequence.
Sample Input
4 4 1 2 3 4 4 2 4 3 1 4 1 1 1 1 5 3 2 1 4 1
Sample Output
1 4 1 2 3 4 2 3 1 2 3 1 4 1 4 1 2 3 4 3 2 1 4 1 2 2 3 5
题意：序列 S={s1,s2,...,sn} 被称为 heapable ，当且仅当存在一个二叉树 T ， n 个点均作为 T 的一个节点，且满足任意非根节点 si 和其父节点 sj , sj≤si 且 j<i 。
现在有一个序列 a1,a2,...,an ，相应将其分成若干个 heapable 的子序列，问使得子序列个数最少的策略
题解：参考自别人的代码 贪心+二分 用memset会TLE 直接for一遍清零。
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define ll __int64 4 #pragma comment(linker, "/STACK:102400000,102400000") 5 int a[100005]; 6 struct node 7 { 8 int f,s; 9 }p; 10 bool operator <(node x,node y) 11 { 12 if(a[x.s]==a[y.s]) return x.s<y.s; 13 return a[x.s]<a[y.s]; 14 } 15 set<node> st; 16 set<node> :: iterator it; 17 vector<int> v[100005]; 18 int dis[100005]; 19 int main() 20 { 21 int t; 22 scanf("%d",&t); 23 for(int i=1;i<=t;i++) 24 { 25 st.clear(); 26 int cnt=0; 27 int n; 28 scanf("%d",&n); 29 for(int k=0;k<=n;k++) 30 dis[k]=0; 31 for(int j=1;j<=n;j++) 32 { 33 scanf("%d",&a[j]); 34 p.s=j; 35 it=st.upper_bound(p); 36 if(it==st.begin()) 37 { 38 v[cnt].push_back(j); 39 p.f=cnt; 40 p.s=j; 41 st.insert(p); 42 cnt++; 43 } 44 else 45 { 46 it--; 47 p=*it; 48 dis[p.s]++; 49 if(dis[p.s]==2) 50 st.erase(it); 51 p.s=j; 52 st.insert(p); 53 v[p.f].push_back(j); 54 } 55 } 56 printf("%d ",cnt); 57 for(int j=0;j<cnt;j++) 58 { 59 printf("%d",v[j].size()); 60 for(int k=0;k<v[j].size();k++) 61 printf(" %d",v[j][k]); 62 printf(" "); 63 v[j].clear(); 64 } 65 } 66 return 0; 67 }