Submission #1180891

#	Time	Username	Problem	Language	Result	Execution time	Memory
1180891		iamhereforfun	Growing Vegetable is Fun 3 (JOI19_ho_t3)	C++20	0 / 100	0 ms	328 KiB

joi2019_ho_t3

// IamHereForFun //

#include <bits/stdc++.h>

using namespace std;

#define LSOne(S) ((S) & -(S))

const int N = 4e2 + 5;
const int M = 2e5 + 5;
const int Q = 1e5 + 5;
const int LG = 20;
const long long INF = -1e18;
const int MOD = 1e9 + 7;
const int B = 350;

int n, c1, c2, c3, dp1[3][N], dp2[3][N], mn = 1e9, l1[3][3], l2[3][3];
string s;

void solve()
{
    cin >> n >> s;
    c1 = count(s.begin(), s.end(), 'R');
    c2 = count(s.begin(), s.end(), 'G');
    c3 = count(s.begin(), s.end(), 'Y');
    if (max(max(c1, c2), c3) > (n + 1) / 2)
    {
        cout << -1;
        return;
    }
    for (int x = 0; x < 3; x++)
    {
        for (int y = 0; y <= n + 1; y++)
        {
            dp2[x][y] = 1e9;
            dp1[x][y] = 1e9;
        }
        for(int y = 0; y < 3; y++)
        {
            l1[x][y] = 1e9;
            l2[x][y] = 1e9;
        }
    }
    // th chi co 2 cay
    // |cnt1 - cnt2| <= 1
    // => zigzag, sau do tinh dc

    // th chinh N <= 60?
    // dp[i][cntR][cntG] = so op it nhat de lượng cây ở sau bang cntR?
    // dp[3][i][cnt]? số op ít nhất để lượng cây có tính chat như trên =))
    // dung dp1, dp2 để op bộ nhớ + dễ code
    // th1: s1 = 'R'
    // cnt > 1
    // dp1[0][cnt] = dp2[0][cnt - 1]
    // dp1[1][cnt] = dp2[1][cnt + 1] + cnt
    // dp1[2][cnt] = dp2[2][cnt + 1] + cnt
    for (int x = 0; x < n; x++)
    {
        int i = 0;
        char c = s[x];
        if (c == 'R')
            i = 0;
        if (c == 'G')
            i = 1;
        if (c == 'Y')
            i = 2;
        if (x == 0)
        {
            for(int y = 0; y < 3; y++) if(y != i) l1[y][i] = 0;
            continue;
        }
        for (int y = 0; y < 3; y++)
        {
            for (int z = 2; z <= x + 1; z++)
            {
                if (y == i)
                {
                    dp2[y][z] = min(dp2[y][z], dp1[y][z - 1]);
                    if (z == 2)
                    {
                        for (int t = 0; t < 3; t++)
                        {
                            dp2[y][z] = min(dp2[y][z], l1[t][y]);
                        }
                    }
                }
                else
                {
                    dp2[y][z] = min(dp2[y][z], dp1[y][z + 1] + z);
                }
            }
            if (y == i)
            {
                // y == current color
                for (int z = 0; z < 3; z++)
                {
                    if (z != y)
                    {
                        for (int t = 0; t < 3; t++)
                        {
                            l2[z][y] = min(l2[z][y], l1[t][z]);
                        }
                    }
                }
            }
            else
            {
                // y != current color
                l2[i][y] = min(l2[i][y], dp1[y][2] + 1);
                for(int z = 0; z < 3; z++)
                {
                    if(z != i)
                    {
                        l2[i][y] = min(l2[i][y], l1[z][y] + 1);
                    }
                }
            }
        }
        for (int y = 0; y < 3; y++)
        {
            for (int z = 1; z <= x + 1; z++)
            {
                dp1[y][z] = dp2[y][z];
                // cout << dp1[y][z] << " " << y << " " << z << " " << x << "\n";
                dp2[y][z] = 1e9;
            }
            for(int z = 0; z < 3; z++)
            {
                l1[y][z] = l2[y][z];
                l2[y][z] = 1e9;
            }
        }
    }
    for (int x = 0; x < 3; x++)
    {
        for(int y = 0; y < 3; y++)
        {
            mn = min(mn, l1[x][y]);
        }
    }
    cout << mn;
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    // cin >> t;
    for (int x = 1; x <= t; x++)
    {
        solve();
    }
    return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 55
• Subtask 3: 0 / 15
• Subtask 4: 0 / 25