| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1057226 | Faisal_Saqib | 팀들 (IOI15_teams) | C++17 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
struct Node {
int lc = 0, rc = 0;
int sum = 0;
};
unordered_map<int, Node> seg;
int nfi = 1; // Start node ID from 1
int build(int l, int r) {
int cur = nfi++;
if (l == r) return cur;
int mid = (l + r) / 2;
seg[cur].lc = build(l, mid);
seg[cur].rc = build(mid + 1, r);
return cur;
}
int update(int node, int l, int r, int x) {
int cur = nfi++;
seg[cur] = seg[node];
if (l == r) {
seg[cur].sum++;
return cur;
}
int mid = (l + r) / 2;
if (x <= mid) seg[cur].lc = update(seg[node].lc, l, mid, x);
else seg[cur].rc = update(seg[node].rc, mid + 1, r, x);
seg[cur].sum = seg[seg[cur].lc].sum + seg[seg[cur].rc].sum;
return cur;
}
int get(int node, int l, int r, int ql, int qr) {
if (ql <= l && r <= qr) return seg[node].sum;
int mid = (l + r) / 2;
if (qr <= mid) return get(seg[node].lc, l, mid, ql, qr);
else if (mid < ql) return get(seg[node].rc, mid + 1, r, ql, qr);
else return get(seg[node].lc, l, mid, ql, qr) + get(seg[node].rc, mid + 1, r, ql, qr);
}
int n, till[N], cup;
vector<int> add[N];
void init(int Ng, int A[], int B[]) {
n = Ng;
for (int i = 0; i < n; i++) {
add[A[i]].push_back(B[i]);
}
till[0] = build(1, n);
cup = till[0];
for (int i = 1; i <= n; i++) {
for (auto& j : add[i])
cup = update(cup, 1, n, j);
till[i] = cup;
}
}
int Contains(int l, int r) {
return get(till[l], 1, n, r, n);
}
int cnt[N]; // cnt of x
int dp[N]; // dp
int can(int m, int k[]) {
int sm = 0;
for (int i = 0; i < m; i++) {
sm += k[i];
if (sm > n) {
// Not Possible
return 0;
}
}
for (int i = 0; i < m; i++)
cnt[k[i]]++;
sort(k, k + m);
int mp = unique(k, k + m) - k;
// Here we write the dp solution
int mi = 1e9;
for (int i = 0; i < mp; i++) {
int x = k[i];
int sz = cnt[x] * x;
int cntx = Contains(x, x);
dp[i] = cntx - sz;
for (int j = i - 1; j >= 0; j--)
dp[i] = min(dp[i], dp[j] + cntx - Contains(k[j], x) - sz);
mi = min(mi, dp[i]);
}
for (int i = 0; i < mp; i++)
cnt[k[i]] = 0;
return (mi >= 0);
}