QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#77177 | #5507. Investors | UCSC_Ravioli# | WA | 6ms | 3708kb | C++20 | 2.3kb | 2023-02-13 08:02:43 | 2023-02-13 08:02:46 |
Judging History
answer
// qdd on Feb 12, 2023
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define ALL(x) begin(x), end(x)
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
for (T& x : v) is >> x;
return is;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
bool f = 0;
for (const T& x : v) (f ? os << ' ' : os) << x, f = 1;
return os;
}
// repeating elements with same id
template <class T>
vector<int> unique_dc(const vector<T>& a, int start_id) {
int n = a.size();
vector<T> t(a);
sort(t.begin(), t.end());
t.resize(unique(t.begin(), t.end()) - t.begin());
vector<int> id(n);
for (int i = 0; i < n; i++) {
id[i] = start_id + lower_bound(t.begin(), t.end(), a[i]) - t.begin();
}
return id;
}
struct fenwick {
int n;
vector<int> t;
fenwick(int n) : n(n), t(n + 1) {}
void add(int p, int x) {
for (; p <= n; p += p & -p) t[p] += x;
}
int get(int p) {
int a = 0;
for (; p > 0; p -= p & -p) a += t[p];
return a;
}
};
int inv[6001][6001];
void sol() {
int n, k;
cin >> n >> k;
vector<int> a(n);
cin >> a;
a = unique_dc(a, 1);
int mx = *max_element(ALL(a));
// inv[i][j] => invs in [i, j]
for (int i = 0; i < n; i++) {
fenwick t(mx);
int now = 0;
for (int j = i; j < n; j++) {
now += t.get(mx) - t.get(a[j]);
t.add(a[i], 1);
inv[i][j] = now;
}
}
vector<ll> dp_before(n), dp_cur(n);
// compute dp_cur[l], ... dp_cur[r] (inclusive)
function<void(int, int, int, int)> compute = [&](int l, int r, int optl, int optr) {
if (l > r) return;
int mid = (l + r) >> 1;
pair<ll, int> best = {LLONG_MAX, -1};
for (int j = optl; j <= min(mid, optr); j++) {
best = min(best, {(j ? dp_before[j - 1] : 0) + inv[j][mid], j});
}
dp_cur[mid] = best.first;
int opt = best.second;
compute(l, mid - 1, optl, opt);
compute(mid + 1, r, opt, optr);
};
for (int i = 0; i < n; i++) {
dp_before[i] = inv[0][i];
}
for (int i = 1; i <= k; i++) {
compute(0, n - 1, 0, n - 1);
dp_before = dp_cur;
}
cout << dp_before[n - 1] << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T;
cin >> T;
while (T--) {
sol();
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3472kb
input:
2 6 1 4 5 6 2 2 1 6 2 4 5 6 2 2 1
output:
2 0
result:
ok 2 lines
Test #2:
score: -100
Wrong Answer
time: 6ms
memory: 3708kb
input:
349 6 2 2 1 2 1 2 1 48 12 42 47 39 39 27 25 22 44 45 13 5 48 38 4 37 6 24 10 42 38 12 37 31 19 44 6 29 17 7 12 7 26 35 24 15 9 37 3 27 21 33 29 34 20 14 30 31 21 48 12 1 43 17 46 17 39 40 22 25 2 22 12 4 11 38 12 4 11 1 5 39 44 37 10 19 20 42 45 2 45 37 20 48 34 16 41 23 18 13 44 47 21 29 4 23 18 16...
output:
0 0 0 15 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 5 4 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 53 0 0 0 6 0 0 0 0 3 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ...
result:
wrong answer 1st lines differ - expected: '1', found: '0'