Submission #1055218

#	Time	Username	Problem	Language	Result	Execution time	Memory
1055218		j_vdd16	Copy and Paste 3 (JOI22_copypaste3)	C++17	15 / 100	3122 ms	917276 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;

int n;
string s;
int A, B, C;
int maxCost;

vector<vvb> isEqualSubstring;
vvi substringCount;
map<pair<ii, ii>, int> memoize;
int solve(int a, int b, int c, int d) {
    if (substringCount[c][d] <= 1 && d > c) return maxCost;

    if (a == b) {
        a = 0;
        b = 0;
    }
    if (c == d) {
        c = 0;
        d = 0;
    }

    if (d - c == 0) return (b - a) * A;

    if (memoize.count({{a, b}, {c, d}})) 
        return memoize[{{a, b}, {c, d}}];

    /*
    [a, b, c, d] depends on:
    [a, x < b, c, d]
    
    [0, 0, c, d] depends on:
    [c, d, x < c, x <= y < d] && y - x < d - c
    
    */

    int res = maxCost;
    if (b - 1 >= a)
        res = min(res, solve(a, b - 1, c, d) + A);
    
    if (b - (d - c) >= a && d > c && isEqualSubstring[c][d][b - (d - c)]) {
        res = min(res, solve(a, b - (d - c), c, d) + C);
    }
    
    if (a == 0 && b == 0) {
        // loop(x, c) {
        //     for (int y = x; y < x + d - c; y++) { //y - x < d - c
        //         res = min(res, solve(c, d, x, y) + B);
        //     }
        // }
        for (int x = c; x <= d; x++) {
            for (int y = x; y < d; y++) { //y - x < d - c
                res = min(res, solve(c, d, x, y) + B);
            }
        }
    }

    memoize[{{a, b}, {c, d}}] = res;
    return res;
}

vector<vector<vvi>> dp;
int& get(int a, int b, int c, int d) {
    if (a == b) {
        a = 0;
        b = 0;
    }
    if (c == d) {
        c = 0;
        d = 0;
    }

    return dp[a][b][c][d];
}

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

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

    cin >> n;

    cin >> s;

    cin >> A >> B >> C;

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

    isEqualSubstring = vector<vvb>(n + 1, vvb(n + 1, vb(n + 1)));
    loop(i, n) {
        loop(j, n) {
            int sz = 1;
            while (max(i, j) + sz <= n && s[i + sz - 1] == s[j + sz - 1]) {
                isEqualSubstring[i][i + sz][j] = true;
                sz++;
            }
        }
    }

    substringCount = vvi(n + 1, vi(n + 1));
    loop(i, n + 1) {
        for (int j = i + 1; j < n + 1; j++) {
            int start = 0;
            while (start < n) {
                if (isEqualSubstring[i][j][start]) {
                    start += j - i;
                    substringCount[i][j]++;
                }
                else {
                    start++;
                }
            }
        }
    }
    
    int result = maxCost;
    loop(c, n + 1) {
        for (int d = c; d < n + 1; d++) {
            //result = min(result, dp[0][n][c][d]);
            result = min(result, solve(0, n, c, d));
        }
    }

    cout << result << endl;

    /*
    dp[a][b][c][d]
    */
    // loop(i, n) {
    //     dp[i][i][n][n] = A;
    // }
    // for (int i = 1; i <= 2 * n; i++) {

    // }

    /*
    dp[a][b][c][d]

    dp[a][b][c][d] = min(dp[a][b][c][d], dp[a][b - 1][c][d] + A)
    dp[0][0][c][d] = min(dp[c][d][a][b] + B)

    if (s[b-(d-c), b] == s[c, d]) dp[a][b][c][d] = min(dp[a][b][c][d], dp[a][b-(d-c)][c][d] + C)

    */
    // dp = vector<vector<vvi>>(n + 1, vector<vvi>(n + 1, vvi(n + 1, vi(n + 1, maxCost))));

    // dp[0][0][0][0] = 0;
    // loop(c, n + 1) {
    //     for (int d = c; d < n + 1; d++) {
    //         loop(a, n + 1) {
    //             for (int b = a; b < n + 1; b++) {
    //                 /*
    //                 [a, b, c, d] depends on:
    //                 [a, x < b, c, d]
                    
    //                 [0, 0, a, b] depends on:
    //                 [a, b, c, y <= b]
                    
    //                 */

    //                 if (b - 1 >= a)
    //                     get(a, b, c, d) = min(get(a, b, c, d), get(a, b - 1, c, d) + A);
                    
    //                 if (b - (d - c) >= a && d > c && s.substr(b - (d - c), d - c) == s.substr(c, d - c)) {
    //                     get(a, b, c, d) = min(get(a, b, c, d), get(a, b - (d - c), c, d) + C);
    //                 }
                    
    //                 if (d <= b)
    //                     get(0, 0, a, b) = min(get(0, 0, a, b), get(a, b, c, d) + B);
    //             }
    //         }
    //     }
    // }

    return 0;

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