이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define lli long long int
#define mp make_pair
#define eb emplace_back
#define pb push_back
#define X first
#define Y second
#define pii pair<int, int>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
void abc() {cout << endl;}
template <typename T, typename ...U> void abc(T i, U ...j) {
cout << i << ' ', abc(j...);
}
template <typename T> void printv(T l, T r) {
for (; l != r; ++l)
cout << *l << " \n"[l + 1 == r];
}
#define test(x...) abc("[" + string(#x) + "]", x);
const int N = 200000;
int main () {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
int C = 1000, K = 8;
if (n <= 5000)
C = 5000, K = 20;
vector <int> a(n), k(n);
for (int i = 0; i < n; ++i)
cin >> a[i];
for (int i = 0; i < n; ++i)
cin >> k[i], k[i] = min(k[i], K);
vector <int> dp(n, 1), bck(n, -1);
vector <vector <int>> tmp(1 << K, vector <int> (K + 1, -1));
auto upd = [&](int &i, int j) {
if (i == -1)
i = j;
else if (dp[i] < dp[j])
i = j;
};
auto rebuild = [&](int to) {
tmp = vector <vector <int>> (1 << K, vector <int> (K + 1, -1));
for (int i = 0; i < to; ++i) {
int x = __builtin_popcount(a[i]);
upd(tmp[a[i]][x], i);
}
for (int cnt = 1; cnt <= K; ++cnt) {
for (int i = 0; i < K; ++i) {
for (int s = 0; s < 1 << K; ++s) {
if (s >> i & 1) {
upd(tmp[s][cnt], tmp[s ^ (1 << i)][cnt - 1]);
} else {
upd(tmp[s][cnt], tmp[s][cnt]);
}
}
}
}
};
for (int i = 0; i < n; i += C) {
int to = min(i + C, n);
for (int j = i; j < to; ++j) {
if (tmp[a[j]][k[j]] != -1 && dp[j] < dp[tmp[a[j]][k[j]]] + 1)
dp[j] = dp[tmp[a[j]][k[j]]] + 1, bck[j] = tmp[a[j]][k[j]];
}
for (int j = i; j < to; ++j) {
for (int ii = i; ii < j; ++ii) {
if (__builtin_popcount(a[ii] & a[j]) == k[j] && dp[j] < dp[ii] + 1)
dp[j] = dp[ii] + 1, bck[j] = ii;
}
}
if (to != n)
rebuild(to);
}
int mx = max_element(all(dp)) - dp.begin();
cout << dp[mx] << endl;
vector <int> ans;
while (mx != -1) {
ans.pb(mx + 1);
mx = bck[mx];
}
reverse(all(ans));
printv(all(ans));
}
/*
5
5 3 5 3 5
10 1 20 1 20
4
1 2 3 4
10 0 1 0
*/
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