Submission #614838

#	Time	Username	Problem	Language	Result	Execution time	Memory
614838		jophyyjh	Jelly Flavours (IOI20_jelly)	C++14	100 / 100	194 ms	684 KiB

jelly

/**
 * 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 jelly_t {
    int a, b;
};

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<jelly_t> values(n);
    for (int i = 0; i < n; i++)
        values[i].a = a[i], values[i].b = b[i];
    std::sort(values.begin(), values.end(),
              [](const jelly_t& j1, const jelly_t& j2) {
                  return j1.a < j2.a;
              });
    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 >= values[k - 1].a)
                dp[k % 2][t].max_f++, dp[k % 2][t].money_A -= values[k - 1].a;
            if (t >= values[k - 1].b) {
                choice_t c = dp[(k - 1) % 2][t - values[k - 1].b];
                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: 11 / 11
• Subtask 2: 24 / 24
• Subtask 3: 9 / 9
• Subtask 4: 10 / 10
• Subtask 5: 14 / 14
• Subtask 6: 32 / 32