Luo Gu [P5283] XOR [dumplings] [stack] Trie

Subject to the effect:

Topic links: https://www.luogu.org/problemnew/show/P5283
given a sequence found $m$ intervals $[l_i,r_i]$ Such that the maximum interval and the XOR.

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Ideas:

Do it first prefix XOR, so the problem is converted to find $m$ of $l,r$ such that $\sum l \ xor\ r$ large as possible.
Notice $a[l]\ xor\ a[r]=a[r]\ xor\ a[l]$ , so that if the elected $l,r$ , then will choose again $r,l$ . So we can ask the number of times $\times 2$ , so that each two successive query is a set of the same $(l,r)(r,l)$ . Divided by 2 final answer on the line.
We maintain a heap, will each $i$ corresponding $a[i]\ xor\ a[j]$ biggest $j$ thrown inside the reactor, each taken interrogation top of the stack, the next large $a[i]\ xor\ a[j]$ thrown into the heap.
for $x$ seeking $Y$ making $a[x]\ xor\ a[y]$ as the first $k$ large apparently can $Trie$ maintenance. Each time the right to use to determine the value to the left and right subtrees.
So you can not use persistent $Trie$ to the completion of this question. time complexity $O(n\log n)$。

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Code:

#include <queue>
#include <cstdio>
#include <string>
#include <iostream>
#define mp make_pair
using namespace std;
typedef long long ll;

const int N=500010,LG=35;
int n,m,tot=1,trie[N*LG][2],cnt[N],size[N*LG];
ll a[N],ans;
priority_queue<pair<ll,int> > q;

ll read()
{
    ll d=0;
    char ch=getchar();
    while (!isdigit(ch)) ch=getchar();
    while (isdigit(ch))
        d=(d<<3)+(d<<1)+(ll)ch-48LL,ch=getchar();
    return d;
}

void insert(ll x)
{
    int p=1;
    for (int i=LG;i>=0;i--)
    {
        int id=(x>>(ll)i)&1;
        if (!trie[p][id]) trie[p][id]=++tot;
        p=trie[p][id];
        size[p]++;
    }
}

ll find(ll x,int k)
{
    int p=1; 
    ll ans=0;
    for (int i=LG;i>=0;i--)
    {
        int id=(x>>(ll)i)&1;
        if (trie[p][id^1]&&size[trie[p][id^1]]>=k)
        {
            ans=(ans<<1)|1;
            p=trie[p][id^1];
        }
        else if (trie[p][id])
        {
            k-=size[trie[p][id^1]];
         	ans<<=1;
         	p=trie[p][id];
        }
    }
    return ans;
}

int main()
{
    scanf("%d%d",&n,&m);
    insert(0); cnt[0]=1;
    for (int i=1;i<=n;i++)
    {
        a[i]=read();
        a[i]^=a[i-1];
        insert(a[i]);
        cnt[i]=1;
    }
    for (int i=0;i<=n;i++)
        q.push(mp(find(a[i],1),i));
    m*=2;
    while (m--)
    {
    	if (!q.size()) break;
        ans+=q.top().first;
        int i=q.top().second;
        q.pop();
        cnt[i]++;
        if (cnt[i]<=n+1)
            q.push(mp(find(a[i],cnt[i]),i));
    }
    cout<<ans/2LL;
    return 0;
}