Submission #1101930

#	Time	Username	Problem	Language	Result	Execution time	Memory
1101930		hqminhuwu	Growing Vegetable is Fun 3 (JOI19_ho_t3)	C++14	0 / 100	143 ms	1048576 KiB

joi2019_ho_t3

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair <ll,ll> pll;
typedef pair <int,int> pii;
typedef pair <int,pii> piii;

#define forr(_a,_b,_c) for(int _a = (_b); _a <= int (_c); ++_a)
#define ford(_a,_b,_c) for(int _a = (_b) + 1; _a --> int (_c);)
#define forf(_a,_b,_c) for(int _a = (_b); _a < int (_c); ++_a)
#define st first
#define nd second
#define pb push_back
#define mp make_pair
#define all(x) begin(x),end(x)
#define mask(i) (1LL << (i))
#define bit(x, i) (((x) >> (i)) & 1)
#define bp __builtin_popcountll
#define file "test"

template<class X, class Y>
	bool minz(X &x, const Y &y) {
		if (x > y) {
			x = y;
			return true;
		} return false;
	}
template<class X, class Y>
	bool maxz(X &x, const Y &y) {
		if (x < y) {
			x = y;
			return true;
		} return false;
	}

const int N = 4e2 + 5;
const ll oo = (ll) 1e16;
const ll mod = 1e9 + 7; // 998244353;

ll n, dp[N][N][N][3], c[3];
string s;

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0);
	#ifdef kaguya
		freopen(file".inp", "r", stdin); freopen(file".out", "w", stdout);
	#endif

	cin >> n >> s;

	s = '#' + s;
	memset (dp, 63, sizeof dp);
	dp[0][0][0][0] = dp[0][0][0][1] = dp[0][0][0][2] = 0;

	forr (i, 1, n){
		if (s[i] == 'R'){
			c[0]++;
		} else if (s[i] == 'G'){
			c[1]++;
		} else {
			c[2]++;
		}

		forf (j, 0, i){
			forr (k, 0, i - j){
				forr (t, 0, 2){
					forr (w, 0, 2){
						if (w == t) continue;
						int nj = j, nk = k;
						
						if (w == 0){
							nj++;
						} else if (w == 1){
							nk++;
						}

						minz(dp[i][nj][nk][w], dp[i - 1][j][k][t]);
					}
				}
			}
		}

		forr (j, 0, i){
			forr (k, 0, i - j){
				forr (t, 0, 2){
					dp[i][j][k][t] += abs(c[0] - j) + abs(c[1] - k) + abs(c[2] - (i - j - k));
				}
			}
		}

		// forr (j, 0, i){
		// 	forr (k, 0, i - j){
		// 		cout << dp[i][j][k][0] << " " << dp[i][j][k][1] << " " << dp[i][j][k][2] << " ";
		// 	}
		// 	cout << "\n";
		// }
		// cout << "\n";
	}

	ll res = oo;

	forr (i, 0, 2)
		minz (res, dp[n][c[0]][c[1]][i]);
	

	cout << (res >= oo ? -1 : res / 2) << "\n";

	return 0;
}
/*



*/

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