Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P5283 [十二省联考 2019] 异或粽子

P5283 [十二省联考 2019] 异或粽子

和超级钢琴一题很像。对于每一个位置作为右端点,选择一个左端点使得值最大,答案一定只在其中出现。开始只有 $n$ 个,每次选择一个后,对原左端点两边继续找作为左端点最大的。我们需要快速找到区间中某一个数和给定一个数异或值最大,可以用可持久化Trie 。对于每个前缀建树,在点上标记编号最大的位置的编号。

查看代码
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#include <cstdio>
#include <queue>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef long long L;
typedef unsigned int U;
const int N = 5e5 + 10, M = 32;
int n, m;
U w[N];
int idx, rt[N];
struct Node { int s[2], v, id; } tr[N * (M + 1)];
void insert (int &x, int k, int id, int t = M - 1)
{
tr[++idx] = tr[x], x = idx;
++tr[x].v, tr[x].id = id;
if (!~t) return;
insert(tr[x].s[k >> t & 1], k, id, t - 1);
}
int query (int x, int y, int k, int t = M - 1)
{
if (!~t) return tr[y].id;
int c = k >> t & 1;
if (tr[tr[y].s[c ^ 1]].v > tr[tr[x].s[c ^ 1]].v)
return query(tr[x].s[c ^ 1], tr[y].s[c ^ 1], k, t - 1);
return query(tr[x].s[c], tr[y].s[c], k, t - 1);
}
struct Data
{
int a, b, l, r;
U calc () const { return w[r] ^ w[l]; }
bool operator < (const Data &_) const
{ return calc() < _.calc(); }
};
int main ()
{
read(n, m);
insert(rt[1], 0, 0);
for (int i = 1; i <= n; ++i)
read(w[i]), insert(rt[i + 1] = rt[i], w[i] ^= w[i - 1], i);
priority_queue <Data> q;
for (int i = 1; i <= n; ++i)
q.push((Data){0, i - 1, query(rt[0], rt[i], w[i]), i});
L res = 0;
while (m--)
{
Data t = q.top(); q.pop();
res += t.calc();
if (t.l < t.b) q.push((Data){t.l + 1, t.b, query(rt[t.l + 1], rt[t.b + 1], w[t.r]), t.r});
if (t.l > t.a) q.push((Data){t.a, t.l - 1, query(rt[t.a], rt[t.l], w[t.r]), t.r});
}
write(res);
return 0;
}