zoj-3963 Heap Partition(贪心+二分+树状数组)

题目链接：

Heap Partition

Time Limit: 2 Seconds Memory Limit: 65536 KB Special Judge
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：给出一个序列，然后要求分成最少多少个子序列，使得每个子序列都满足上面的要求

思路：贪心，对于a[i],贪心的话就是要在a[1]~a[i-1]中找到一个a[j]做父亲（且a[j]不能超过两个孩子）,a[j]<=a[i]&&a[j]>=a[k](1<=任意k<=i-1,k!=j)
　　　可以离散化，然后二分+树状数组找，线段树会T;
AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}

int n,a[maxn],vis[maxn],p[maxn],b[maxn],sum[maxn];
vector<int>ve[maxn];
struct node
{
int a,id;
}po[maxn];
int cmp(node x,node y)
{
if(x.a==y.a)return x.id<y.id;
return x.a<y.a;
}
inline int lowbit(int x){return x&(-x);}
inline int query(int x)
{
int s=0;
while(x)
{
s+=sum[x];
x-=lowbit(x);
}
return s;
}
inline void update(int x,int num)
{
while(x<=n)
{
sum[x]+=num;
x+=lowbit(x);
}
return ;
}

inline int solve(int x)
{
int l=1,r=b[x]-1;
while(l<=r)
{
int mid=(l+r)>>1;
if(query(b[x]-1)-query(mid-1)>0)l=mid+1;
else r=mid-1;
}
if(l-1<=0)return -1;
return p[l-1];
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)read(po[i].a),po[i].id=i,ve[i].clear(),sum[i]=0;
sort(po+1,po+n+1,cmp);
for(int i=1;i<=n;i++)b[po[i].id]=i,p[i]=po[i].id;
int ans=0;
for(int i=1;i<=n;i++)
{
int pos=solve(i);
if(pos==-1)ans++,vis[i]=ans,ve[ans].push_back(i);
else vis[i]=vis[pos],ve[vis[i]].push_back(i),update(b[pos],-1);
update(b[i],2);
}
printf("%d\n",ans);
for(int i=1;i<=ans;i++)
{
int len=ve[i].size();
printf("%d",len);
for(int j=0;j<len;j++)printf(" %d",ve[i][j]);puts("");
}
}
return 0;
}