HDU 5338(ZZX and Permutations-用线段树贪心)

ZZX and Permutations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 888    Accepted Submission(s): 278

Problem Description

ZZX likes permutations.



ZZX knows that a permutation can be decomposed into disjoint cycles(see https://en.wikipedia.org/wiki/Permutation#Cycle_notation). For example:

145632=(1)(35)(462)=(462)(1)(35)=(35)(1)(462)=(246)(1)(53)=(624)(1)(53)……

Note that there are many ways to rewrite it, but they are all equivalent.

A cycle with only one element is also written in the decomposition, like (1) in the example above.



Now, we remove all the parentheses in the decomposition. So the decomposition of 145632 can be 135462,462135,351462,246153,624153……



Now you are given the decomposition of a permutation after removing all the parentheses (itself is also a permutation). You should recover the original permutation. There are many ways to recover, so you should find the one with largest lexicographic order.

 

Input

First line contains an integer
t,
the number of test cases.

Then t
testcases follow. In each testcase:

First line contains an integer n,
the size of the permutation.

Second line contains n
space-separated integers, the decomposition after removing parentheses.



n≤105.
There are 10 testcases satisfying n≤105,
200 testcases satisfying n≤1000.

 

Output

Output n
space-separated numbers in a line for each testcase.

Don't output space after the last number of a line.

 

Sample Input

2
6
1 4 5 6 3 2
2
1 2

 

Sample Output

4 6 2 5 1 3
2 1

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 4

 

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从第1位開始贪心，每位要么取前面的（包含自己）要么取后一个。

每次找到前面能取的最大值。

假设取后面的那么无论，仅仅是以后不能取后一位，

若取前面的则拿走一段

注意反例

6 7 1 2

要用set找出前面没间隔的部分

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<set>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEM2(a,i) memset(a,i,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
class SegmentTree
{
    ll a[MAXN*4],minv[MAXN*4],sumv[MAXN*4],maxv[MAXN*4],addv[MAXN*4],setv[MAXN*4];
    int n;
public:
    SegmentTree(){MEM(a) MEM(minv) MEM(sumv) MEM(maxv) MEM(addv) MEM2(setv,-1) }
    SegmentTree(int _n):n(_n){MEM(a) MEM(minv) MEM(sumv) MEM(maxv) MEM(addv) MEM2(setv,-1) }
    void mem(int _n)
    {
        n=_n;
        MEM(a) MEM(minv)  MEM(sumv) MEM(maxv) MEM(addv) MEM2(setv,-1)
    }  

    void maintain(int o,int L,int R)
    {

		sumv[o]=maxv[o]=minv[o]=0;
    	if (L<R) //仅仅考虑左右子树
		{
			sumv[o]=sumv[Lson]+sumv[Rson];
			minv[o]=min(minv[Lson],minv[Rson]);
			maxv[o]=max(maxv[Lson],maxv[Rson]);
		} //仅仅考虑add操作
		if (setv[o]>=0) sumv[o]=setv[o]*(R-L+1),minv[o]=maxv[o]=setv[o];

		minv[o]+=addv[o];maxv[o]+=addv[o];sumv[o]+=addv[o]*(R-L+1);
    }

	int y1,y2,v;
	void update(int o,int L,int R) //y1,y2,v
	{
		if (y1<=L&&R<=y2) {
			addv[o]+=v;
		}
		else{
			pushdown(o);
			int M=(R+L)>>1;
			if (y1<=M) update(Lson,L,M); else maintain(Lson,L,M);
			if (M< y2) update(Rson,M+1,R); else maintain(Rson,M+1,R);
		}

		maintain(o,L,R); 

	}
	void update2(int o,int L,int R)
	{
		if (y1<=L&&R<=y2) {
			setv[o]=v;addv[o]=0;
		}
		else{
			pushdown(o);
			int M=(R+L)>>1;
			if (y1<=M) update2(Lson,L,M); else maintain(Lson,L,M); //维护pushodown，再次maintain
			if (M< y2) update2(Rson,M+1,R); else maintain(Rson,M+1,R);
		}

		maintain(o,L,R);
	}

	void pushdown(int o)
	{
		if (setv[o]>=0)
		{
			setv[Lson]=setv[Rson]=setv[o];
			addv[Lson]=addv[Rson]=0;
			setv[o]=-1;
		}
		if (addv[o])
		{
			addv[Lson]+=addv[o];
			addv[Rson]+=addv[o];
			addv[o]=0;
		}
	}
	ll _min,_max,_sum; 

	void query2(int o,int L,int R,ll add)
	{
		if (setv[o]>=0)
		{
			_sum+=(setv[o]+addv[o]+add)*(min(R,y2)-max(L,y1)+1);
			_min=min(_min,setv[o]+addv[o]+add);
			_max=max(_max,setv[o]+addv[o]+add);
		} else if (y1<=L&&R<=y2)
		{
			_sum+=sumv[o]+add*(R-L+1);
			_min=min(_min,minv[o]+add);
			_max=max(_max,maxv[o]+add);
		} else {
		//	pushdown(o);
			int M=(L+R)>>1;
			if (y1<=M) query2(Lson,L,M,add+addv[o]);// else maintain(Lson,L,M);
			if (M< y2) query2(Rson,M+1,R,add+addv[o]);// else maintain(Rson,M+1,R);
		}
		//maintain(o,L,R);
	}

	void query(int o,int L,int R,ll add) //y1,y2
	{
		if (y1<=L&&R<=y2)
		{
			_sum+=sumv[o]+add*(R-L+1);
			_min=min(_min,minv[o]+add);
			_max=max(_max,maxv[o]+add);
		}
		else{
			int M=(R+L)>>1;
			if (y1<=M) query(Lson,L,M,add+addv[o]);
			if (M< y2) query(Rson,M+1,R,add+addv[o]);
		}
	}

	void add(int l,int r,ll v)
	{
		y1=l,y2=r;this->v=v;
		update(1,1,n);
	}
	void set(int l,int r,ll v)
	{
		y1=l,y2=r;this->v=v;
		update2(1,1,n);
	}
	ll ask(int l,int r,int b=0)
	{
		_sum=0,_min=INF,_max=-1;
		y1=l,y2=r;
		query2(1,1,n,0);
	//	cout<<_sum<<' '<<_max<<' '<<_min<<endl; 

		switch(b)
		{
			case 1:return _sum;
			case 2:return _min;
			case 3:return _max;
			default:break;
		}
	}
	void print()
	{
		For(i,n)
			cout<<ask(i,i,1)<<' ';
		cout<<endl;

	}

    //先set后add
}S;
int h[MAXN],n,a[MAXN];
int b[MAXN],ans[MAXN];
set<int> S2;
set<int>::iterator it;

int main()
{
//	freopen("L.in","r",stdin);

	int T;cin>>T;
	while(T--) {
		cin>>n;
		For(i,n) b[i]=1;
		S.mem(n);
		S2.clear();S2.insert(1);
		For(i,n) scanf("%d",&a[i]),S.add(i,i,a[i]);
		a[0]=a[n+1]=0;
		For(i,n) h[a[i]]=i;

		For(i,n) {
			int t=h[i];
			if (!b[t]) continue;

			it=S2.upper_bound(t);
			it--;
			int l=(*it);
			ll premax = S.ask(l,t,3);

			if (premax>a[t+1]*b[t+1]) {
				int t2=h[premax];
				Fork(j,t2,t) b[j]=0;
				Fork(j,t2,t-1) ans[a[j]]=a[j+1]; ans[a[t]]=a[t2]; 

				S.set(t,t2,0);
				S2.insert(t+1);

			} else {
				S.set(t+1,t+1,0);
			}

		}

		For(i,n-1) printf("%d ",ans[i]);
		printf("%d\n",ans[n]); 

	}

	return 0;
}