oj.uz

Submission #614829

# Submission time Handle Problem Language Result Execution time Memory
614829 2022-07-31T05:05:38 Z jophyyjh Jelly Flavours (IOI20_jelly) C++14
0 / 100
 7 ms 468 KB 
/**
 * A really nice dp + greedy problem~ I can recall that this is one the first few IOI
 * problems I ever saw, and I also heard the solution from tinjyu.
 * 
 * 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"

typedef std::vector<int>	vec;

const int INF = 0x3f3f3f3f;


struct choice_t {
    int max_f, money_A;
};

inline bool operator<(const choice_t& c1, const choice_t& c2) {
    return c1.max_f < c2.max_f || (c1.max_f == c2.max_f && c1.money_A < c2.money_A);
}

int find_maximum_unique(int x, int y, vec a, vec b) {
    int n = a.size();
    std::vector<std::vector<choice_t>> dp(2, std::vector<choice_t>(y + 1));
    dp[0][0] = choice_t{0, x};
    for (int t = 1; t <= y; t++)
        dp[0][t] = choice_t{-INF, -INF};
    
    int ans = 0;
    for (int k = 1; k <= n; k++) {
        for (int t = 0; t <= y; t++) {
            dp[k % 2][t] = dp[(k - 1) % 2][t];
            if (dp[k % 2][t].money_A >= a[k - 1])
                dp[k % 2][t].max_f++, dp[k % 2][t].money_A -= a[k - 1];
            if (t >= b[k - 1]) {
                choice_t c = dp[(k - 1) % 2][t - b[k - 1]];
                c.max_f++;
                dp[k % 2][t] = std::max(dp[k % 2][t], c);
            }
            ans = std::max(ans, dp[k % 2][t].max_f);
        }
    }
    return ans;
}

Subtask #1
0 / 11.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 212 KB Output is correct
2 Incorrect 1 ms 212 KB 1st lines differ - on the 1st token, expected: '7', found: '6'
3 Halted 0 ms 0 KB -

Subtask #2
0 / 24.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 212 KB Output is correct
2 Incorrect 1 ms 212 KB 1st lines differ - on the 1st token, expected: '7', found: '6'
3 Halted 0 ms 0 KB -

Subtask #3
0 / 9.0

# Verdict Execution time Memory Grader output
1 Incorrect 1 ms 340 KB 1st lines differ - on the 1st token, expected: '689', found: '648'
2 Halted 0 ms 0 KB -

Subtask #4
0 / 10.0

# Verdict Execution time Memory Grader output
1 Incorrect 7 ms 340 KB 1st lines differ - on the 1st token, expected: '62', found: '12'
2 Halted 0 ms 0 KB -

Subtask #5
0 / 14.0

# Verdict Execution time Memory Grader output
1 Incorrect 5 ms 468 KB 1st lines differ - on the 1st token, expected: '154', found: '63'
2 Halted 0 ms 0 KB -

Subtask #6
0 / 32.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 212 KB Output is correct
2 Incorrect 1 ms 212 KB 1st lines differ - on the 1st token, expected: '7', found: '6'
3 Halted 0 ms 0 KB -