Submission #765572

#	Time	Username	Problem	Language	Result	Execution time	Memory
765572		NK_	Growing Vegetable is Fun 3 (JOI19_ho_t3)	C++17	100 / 100	268 ms	2712 KiB

joi2019_ho_t3

// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>

using namespace std;

#define nl '\n'
#define pb push_back
#define sz(x) int(x.size())

using str = string;
template<class T> using V = vector<T>;

const int INF = int(1e9) + 10;

int main() {
	cin.tie(0)->sync_with_stdio(0);
	
	int N; cin >> N;
	str S; cin >> S;

	V<V<int>> oc(3); for(int i = 0; i < N; i++) {
		int t = (S[i] == 'R' ? 0 : (S[i] == 'G' ? 1 : 2));
		oc[t].pb(i);
	}

	sort(begin(oc), end(oc), [&](const V<int> &a, const V<int> &b) {
		return size(a) < size(b);
	});


	int A = sz(oc[0]), B = sz(oc[1]), C = N - A - B;

	V<V<V<int>>> COST(9);
	for(int i = 0; i < 3; i++) for(int j = 0; j < 3; j++) if (i != j) {
		COST[3 * i + j] = V<V<int>>(sz(oc[i]), V<int>(sz(oc[j]) + 1));

		for(int x = 0; x < sz(oc[i]); x++) for(int y = 0; y <= sz(oc[j]); y++) {
			COST[3 * i + j][x][y] = (lower_bound(begin(oc[j]) + y, end(oc[j]), oc[i][x]) - (begin(oc[j]) + y));
			// cout << 3 * i + j << " " << x << " " << y << endl;
			// cout << COST[3 * i + j][x][y] << endl;
		}	

	}


	V<V<V<int>>> dp(A + 1, V<V<int>>(B + 1, V<int>(3, INF)));
	dp[0][0] = {0, 0, 0};
	for(int i = 0; i < N; i++) {	
		V<V<V<int>>> ndp(A + 1, V<V<int>>(B + 1, V<int>(3, INF)));

		for(int a = 0; a <= A; a++) for(int b = 0; b <= B; b++) for(int lst = 0; lst < 3; lst++) {
			if (dp[a][b][lst] == INF) continue;
			int c = i - a - b; if (c < 0) continue;
			// cout << i << " " << a << " " << b << " " << c << " " << lst << endl;
			// cout << dp[a][b][lst] << endl;
			// cout << endl;

			if (a + 1 <= A && lst != 0) 
				ndp[a+1][b][0] = min(ndp[a+1][b][0], dp[a][b][lst] + COST[1][a][b] + COST[2][a][c]);

			// cout << 1 << endl;

			if (b + 1 <= B && lst != 1) 
				ndp[a][b+1][1] = min(ndp[a][b+1][1], dp[a][b][lst] + COST[3][b][a] + COST[5][b][c]);

			// cout << 1 << endl;

			if (c + 1 <= C && lst != 2) 
				ndp[a][b][2] = min(ndp[a][b][2], dp[a][b][lst] + COST[6][c][a] + COST[7][c][b]);
		}
		// cout << endl;

		dp.swap(ndp);
	}

	int ans = min({dp[A][B][0], dp[A][B][1], dp[A][B][2]});
	cout << (ans == INF ? -1 : ans) << nl;

    return 0;
}

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