# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
673225 | ThegeekKnight16 | Jelly Flavours (IOI20_jelly) | 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 "jelly.h"
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 10;
const int MAXM = 1e4 + 10;
const int INF = 0x3f3f3f3f;
int dp[MAXN][MAXM];
int dp2[MAXN][MAXM];
int n, a[MAXN], b[MAXN];
vector<pair<int, int> > buga;
void limpar(int n, int m)
{
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
dp[i][j] = 0;
dp2[i][j] = 0;
}
}
}
int find_maximum_unique(int x, int y, vector<int> _a, vector<int> _b) {
n = _a.size();
//if (rand() % 2) {
// swap(x, y);
// swap(_a, _b);
// }
for (int i = 0; i < n; i++)
buga.emplace_back(_a[i], _b[i]);
sort(buga.begin(), buga.end());
for (int i = 1; i <= n; i++) {
a[i] = buga[i - 1].first;
b[i] = buga[i - 1].second;
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= x; j++) {
dp[i][j] = dp[i - 1][j];
if (j >= a[i])
dp[i][j] = max(dp[i][j], dp[i - 1][j - a[i]] + 1);
}
}
for(int i = n, j = x; i >= 1; i--)
{
if (j >= a[i] && dp[i][j] == dp[i-1][j - a[i]] + 1) {
j -= a[i]; b[i] = INF;
}
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= y; j++) {
dp2[i][j] = dp2[i - 1][j];
if (j >= b[i])
dp2[i][j] = max(dp2[i][j], dp2[i - 1][j - b[i]] + 1);
}
}
int resp = (dp[n][x] + dp2[n][y]);
limpar(n, max(x, y));
for (int i = 1; i <= n; i++) {
a[i] = buga[i - 1].first;
b[i] = buga[i - 1].second;
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= y; j++) {
dp[i][j] = dp[i - 1][j];
if (j >= b[i])
dp[i][j] = max(dp[i][j], dp[i - 1][j - b[i]] + 1);
}
}
for(int i = n, j = x; i >= 1; i--)
{
if (j >= a[i] && dp[i][j] == dp[i-1][j - b[i]] + 1) {
j -= b[i]; a[i] = INF;
}
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= x; j++) {
dp2[i][j] = dp2[i - 1][j];
if (j >= a[i])
dp2[i][j] = max(dp2[i][j], dp2[i - 1][j - a[i]] + 1);
}
}
return max(resp, dp2[n][x] + dp[n][y]);
}