跟超级钢琴做法差不多, 就是 [l,r] 中的最大值的位置是 k, 那么次大值的位置必定在 [l,k-1] 和 [k+1,r] 其中之一的思想。
#include <bits/stdc++.h>
typedef long long LL;
using namespace std;
const int N = 5e5, K = 2e5;
LL rd() {
LL x = 0ll;
char c = getchar();
while (c < '0' || c > '9') c = getchar();
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x;
}
int n, k; // 1 << 31
LL A[N + 233];
struct Trie { int c[2], col; Trie() {c[0]=c[1]=col=0;} } a[N * 34 + 2333];
int tot, root[N + 233];
void init() {
int p = root[0] = ++ tot;
for (int i = 31; ~ i; -- i) a[p].c[0] = ++ tot, p = tot;
}
void ins (int id, LL x) {
int p = root[id - 1], q = root[id] = ++ tot;
a[q].col = id;
for (int i = 31; ~ i; -- i) {
int v = (x >> i) & 1;
a[a[q].c[v] = ++ tot].col = id;
a[q].c[v ^ 1] = a[p].c[v ^ 1];
p = a[p].c[v], q = tot;
}
}
pair<LL,int> ask (int l, int r, LL x) {
LL res = 0ll;
int p = root[r];
for (int i = 31; ~ i; -- i) {
int v = (x >> i) & 1;
if (a[p].c[v ^ 1] && a[a[p].c[v ^ 1]].col >= l)
res += (1ll << i), p = a[p].c[v ^ 1];
else
p = a[p].c[v];
}
return make_pair (res, a[p].col);
}
struct sta { int l, r, x; LL v; int pos; sta (int a, int b, int c, LL d, int e) : l(a), r(b), x(c), v(d), pos(e) {} sta() {}
} q[N*2+23];
bool operator < (sta s1, sta s2) { return s1.v < s2.v; }
int tp;
int main()
{
init();
n = rd(), k = rd();
for (int i = 1; i <= n; ++ i) A[i] = A[i - 1] ^ rd(), ins (i, A[i]);
for (int i = 1; i <= n; ++ i) {
pair<LL,int> tmp = ask (0, i - 1, A[i]);
q[++ tp] = sta (0, i - 1, i, tmp.first, tmp.second);
push_heap (q + 1, q + 1 + tp);
}
LL ans = 0ll;
for (int i = 1; i <= k; ++ i) {
sta now = q[1];
pop_heap (q + 1, q + 1 + tp);
-- tp;
ans += now.v;
int x = now.x, l = now.l, r = now.r, k = now.pos;
if (l < k) {
pair<LL,int> tmp = ask (l, k - 1, A[x]);
q[++ tp] = sta (l, k - 1, x, tmp.first, tmp.second);
push_heap (q + 1, q + tp + 1);
}
if (k < r) {
pair<LL,int> tmp = ask (k + 1, r, A[x]);
q[++ tp] = sta (k + 1, r, x, tmp.first, tmp.second);
push_heap (q + 1, q + tp + 1);
}
}
cout << ans;
return 0;
}