"ROI 2018 Day 2" no carry addition

Main idea of the title:

Give binary numbers \ (a_1 \ ldots a_n \), for \ (b_1\ldots b_n \)

Meet \ (a_i\leq b_i \), \ (\ bigoplus b_i=\sum b_i \), where $\ bigoplus $is XOR and

Find the minimum value of \ (\ sum b_i \)

Set the length order as \ (N=\sum len(a_i) \)

\(O(N^2-N^3) \), determine each bit of the answer from high to low

The current bit of enumeration is 0 and the following bit is 1. Greedily determine whether there is a scheme

Check whether an answer is legal:

Dynamically maintain an inverted \ (a_i \) set, considering each position from high to low

1. If the current bit is 0:

If there is a number greater than or equal to this bit in \ (a_i \), it is illegal

2. If the current bit is 1:

2-1. If there are 2 numbers with current bit of 1 in \ (a_i \), it is illegal

2-2. If there is exactly one in \ (a_i \), use this 1 for this \ (a_i \), and remove the highest bit of \ (a_i \) and put it back into the set

2-3. Does not exist, use this \ (1 \) to delete the largest \ (a_i \)

In fact, this greed itself is not efficient

Optimization 1: quickly determine the possible range of the highest answer

Make \ (b = \ Max \ {len (a_i) + i-1 \} \)

Then \ (len(Ans)\in[B,B+1] \)

The upper and lower bounds can be obtained by the greedy simulation above

Optimization 2: fast maintenance \ (a_i \) reverse order

Obviously, in the process of continuous change, the current \ (a_i \) must be a suffix of the original \ (a_i \)

Consider sorting all such suffixes. For convenience, use each highest 1 to represent a legal suffix

Obviously, suffixes of the same length can be sorted according to the position of the next 1 in the suffix

That is, a process similar to cardinality sorting. You can maintain the suffix corresponding to the next \ (1 \) in each suffix

Preprocessing complexity is \ (O(N\log N) \)

At the same time, you can also use the segment tree to quickly maintain the insertion / deletion ranking, get the value of \ (B \), and the complexity of a single operation \ (O(\log N) \)

Optimization 3

Call the \ (I \) satisfying \ (len(a_i)+i-1=B \) as \ (\ text{critical number} \)

Make \ (p\) the smallest \ (\text{critical number}\), that is, the first case 2-1 occurs in the process of greed/ 2-2. Location of

Whether the decision answer is \ (B \) or \ (B+1 \), that is, the decision

Delete the highest bit of \ (a_p \) with the position of \ (len(a_p) \), or delete \ (a_p \) with the position of \ (len(a_p)+1 \)

(\ ([1,p-1] \) must be deleted)

\(\ text{intended solution} \) violent recursion is used to complete the operation of determining each bit

Function Solve(Limit) Limit Is the highest 1 currently available
	Seek B,p
	delete a[1,p-1]
	delete a[p]Highest position
	if B<=Limit and Solve(p-1) then
		ans[len(a[p]),B]=1
		return True
	delete a[p]
	if B+1<=Limit and Solve(p) then 
		ans[len(a[p])+1,B+1]=1
		return True
	else return False
end

As for complexity.. I don't know

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define rep(i,a,b) for(int i=a,i##end=b;i<=i##end;++i)
#define drep(i,a,b) for(int i=a,i##end=b;i>=i##end;--i)
template <class T> inline void cmin(T &a,const T &b){ ((a>b)&&(a=b)); }
template <class T> inline void cmax(T &a,const T &b){ ((a<b)&&(a=b)); }
char IO;
int rd(){
	int s=0;
	while(!isdigit(IO=getchar()));
	do s=(s<<1)+(s<<3)+(IO^'0');
	while(isdigit(IO=getchar()));
	return s;
}

typedef vector <int> V;
const int N=3e5+10,INF=1e9+10;

int n,m,I[N],L;
char s[N];
int fir[N],nxt[N],rk[N],len[N],id[N];
V A[N];

struct Affirmation_Of_My_Existence{
	int s[N<<2],t[N<<2];
	void Down(int p){
		rep(v,p<<1,p<<1|1) t[v]+=t[p],s[v]+=t[p];
		t[p]=0;
	}
	void Upd(int p,int l,int r,int ql,int qr,int x) {
		if(ql>qr) return;
		if(ql<=l && r<=qr) {
			s[p]+=x,t[p]+=x;
			return;
		}
		Down(p);
		int mid=(l+r)>>1;
		if(ql<=mid) Upd(p<<1,l,mid,ql,qr,x);
		if(qr>mid) Upd(p<<1|1,mid+1,r,ql,qr,x);
		s[p]=max(s[p<<1],s[p<<1|1]);
	}
	void Build(int p,int l,int r){ 
		s[p]=len[id[l]]-INF;
		if(l==r) return;
		int mid=(l+r)>>1;
		Build(p<<1,l,mid),Build(p<<1|1,mid+1,r);
	}
	void Add(int x,int k){
		x=rk[x];
		Upd(1,1,m,x,x,INF*k),Upd(1,1,m,x+1,m,k);
	}
	// Find the first critical position "p", and return all the bits in [1,p]
	void Get(int p,int l,int r,int x,V &R){
		if(s[p]<0) return;
		if(l==r) return R.pb(id[l]);
		Down(p);
		int mid=(l+r)>>1;
		Get(p<<1,l,mid,x,R);
		if(s[p<<1]!=x) Get(p<<1|1,mid+1,r,x,R);
	}
} T;

int Solve(int L){
	// L denotes the maxmium bit we can use
	int B=T.s[1];
	if(B<0) return 1;
	if(B>L) return 0;
	V R; T.Get(1,1,m,B,R);
	int p=*R.rbegin(),l=len[p];
	for(int i:R) T.Add(i,-1);
	
	// Try ans B , so we use bit [nxt,B] to delete the number [1,p-1] 
    // and the number a[p] will be set to a[p]-2^l
	if(nxt[p]) T.Add(nxt[p],1);
	if(Solve(l-1)) {
		rep(i,l,B) s[i]=1;
		return B+1;
	}

	// Try ans B+1 , so we use bit [nxt+1,B+1] to delete the [1,p]
	if(nxt[p]) T.Add(nxt[p],-1);
	if(B<L && Solve(l)) {
		rep(i,l+1,B+1) s[i]=1;
		return B+2;
	}
	for(int i:R) T.Add(i,1);
	return 0;
}

int main(){
	rep(i,1,n=rd()) {
		scanf("%s",s); int l=strlen(s);
		cmax(L,l);
		drep(j,l-1,0) if(s[j]=='1') {
			nxt[++m]=fir[i];
			A[len[m]=l-j-1].pb(m);
			fir[i]=m;
		}
	}
	rk[0]=1e9;
	int k=m;
	rep(i,0,L-1) {
		k-=A[i].size();
		sort(A[i].begin(),A[i].end(),[&](int x,int y){ return rk[nxt[x]]<rk[nxt[y]]; });
		for(int j:A[i]) id[rk[j]=++k]=j;
		k-=A[i].size();
	}
	T.Build(1,1,m);
	rep(i,1,n) T.Add(fir[i],1);
	memset(s,0,sizeof s);
	drep(i,Solve(INF)-1,0) putchar(s[i]^48);
}