# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
1039514 | Hacv16 | 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.
/*
* Anyway, the first few subtasks are pretty easy. In particular, when there's only
* one person (e.g. when y=0) we can simply sort all flavours according to price (for
* that person) and greedily pick from the small ones. From the constraints, one may
* guess that the only way to AC this problem is to do a O(nx) dp (not a O(nxy) one).
* Now, let dp[i][t] be the dp state when we only consider the first i flavours and
* exactly t dollars are spent on store B. Each dp state already captures the cost in
* B, so it may (?!) natural to maximize the num of unique flavours she can buy while
* minimizing the amount of money spent on store A.
* So, we first SORT the flavours based on their price in store A (from small to
* large). We know that if we've selected a subset of jelly in B, she would just
* greedily pick the remaining ones in store A (from left to right). Therefore, in
* each dp state, we store the max num of flavours we can buy. In addition, among all
* choices that can reach the max. num of flavours, we store the max amount of money
* that can be left with in store A. It's not difficult to prove that this is indeed
* a valid approach using greedy arguments. Nice problem!
*
* Time Complexity: O(ny) (or equivalently, O(nx))
* Implementation 1
*/
#include <bits/stdc++.h>
#include "jelly.h"
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define fr first
#define sc second
const int MAXN = 2005;
const int MAXM = 20050
const int INF = 0x3f3f3f3f;
int a[MAX], b[MAX];
pair<int, int> dp[MAX][MAXM];
int find_maximum_unique(int x, int y, vector<int> _a, vector<int> _b)
{
int n = _a.size();
vector<pair<int, int>> aux;
for(int i = 0; i < n; i++)
aux.emplace_back(_a[i], _b[i]);
sort(aux.begin(), aux.end());
for(int i = 1; i <= n; i++){
a[i] = aux[i - 1].fr;
b[i] = aux[i - 1].sc;
}
dp[0][0] = {0, x};
for(int i = 1; i <= y; i++)
dp[0][i] = {-INF, -INF};
int ans = 0;
for(int i = 1; i <= n; i++){
for(int j = 0; j <= y; j++){
dp[i][j] = dp[i - 1][j];
if(dp[i][j].sc >= a[i])
dp[i][j].fr++, dp[i][j].sc -= a[i];
if(j >= b[i]){
pair<int, int> t = dp[i - 1][j - b[i]];
t.fr++;
if(t.fr > dp[i][j].fr || (t.fr == dp[i][j].fr && t.sc > dp[i][j].sc))
dp[i][j] = t;
}
ans = max(ans, dp[i][j].fr);
}
}
return ans;
}