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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
Recommend wange2014 | We have carefully selected several similar problems for you: 5416

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從第1位開始貪心，每位要麽取前面的（包含自己）要麽取後一個。

每次找到前面能取的最大值。

假設取後面的那麽無論，僅僅是以後不能取後一位，

若取前面的則拿走一段

註意反例

6 7 1 2

要用set找出前面沒間隔的部分

#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) //僅僅考慮左右子樹 
		{
			sumv[o]=sumv[Lson]+sumv[Rson];
			minv[o]=min(minv[Lson],minv[Rson]);
			maxv[o]=max(maxv[Lson],maxv[Rson]);
		} //僅僅考慮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); //維護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後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;
}

HDU 5338(ZZX and Permutations-用線段樹貪心)