Submission #1054599

#	Time	Username	Problem	Language	Result	Execution time	Memory
1054599		j_vdd16	Copy and Paste 3 (JOI22_copypaste3)	C++17	5 / 100	48 ms	49500 KiB

copypaste3

#include <algorithm>
#include <bitset>
#include <cstdint>
#include <cstring>
#include <iostream>
#include <limits.h>
#include <math.h>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define int long long
#define loop(X, N) for(int X = 0; X < (N); X++)
#define all(V) V.begin(), V.end()
#define rall(V) V.rbegin(), V.rend()

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;

typedef uint64_t u64;
typedef int64_t i64;

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);

    //dp cost of building S[a, b] with S[c, d] in clipboard

    int n;
    cin >> n;

    string s;
    cin >> s;

    int A, B, C;
    cin >> A >> B >> C;

    int maxCost = max({A, B, C}) * n;

    // vector<vector<vvi>> dp(n, vector<vvi>(n, vvi(n + 1, vi(n + 1, max({A, B, C}) * n))));
    // loop(i, n) {
    //     dp[i][i][n][n] = A;
    // }
    // for (int i = 1; i <= 2 * n; i++) {

    // }

    // map<string, int> clipBoardCost;
    // clipBoardCost[""] = 0;
    // loop(i, n) {
    //     clipBoardCost[s.substr(i, 1)] = A + B;
    // }
    // for (int sz = 2; sz <= n; sz++) {
    //     loop(i, n) {
    //         string goal = s.substr(i, sz);
    //         string prev = goal;
    //         prev.pop_back();

    //         clipBoardCost[goal] = min(clipBoardCost[goal], sz * A);
            
    //     }
    // }

    vvi cost(n + 1, vi(n + 1, maxCost));
    cost[0][0] = 0;
    
    //cost[0][j] = min(cost[j][i] + B)
    //cost[i][j] = min(cost[i - j][j] + C, cost[i - 1][j] + A)

    loop(i, n + 1) {
        loop(j, n + 1) {
            if (i - 1 >= 0) cost[i][j] = min(cost[i][j], cost[i - 1][j] + A);
            if (i - j >= 0) cost[i][j] = min(cost[i][j], cost[i - j][j] + C);

            cost[0][i] = min(cost[0][i], cost[i][j] + B);
        }
    }

    int result = maxCost;
    loop(j, n + 1) {
        result = min(result, cost[n][j]);
    }

    cout << result << endl;

	return 0;
}

• Subtask 1: 0 / 1
• Subtask 2: 5 / 5
• Subtask 3: 0 / 14
• Subtask 4: 0 / 10
• Subtask 5: 0 / 32
• Subtask 6: 0 / 38