Submission #372629

#	Time	Username	Problem	Language	Result	Execution time	Memory
372629		alishahali1382	Growing Vegetable is Fun 3 (JOI19_ho_t3)	C++14	100 / 100	452 ms	780564 KiB

joi2019_ho_t3

#include <bits/stdc++.h>
#pragma GCC optimize ("O2,unroll-loops")
//#pragma GCC optimize("no-stack-protector,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef pair<ll, ll> pll;
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define debugp(x) cerr<<#x<<"= {"<<(x.first)<<", "<<(x.second)<<"}"<<endl;
#define debug2(x, y) cerr<<"{"<<#x<<", "<<#y<<"} = {"<<(x)<<", "<<(y)<<"}"<<endl;
#define debugv(v) {cerr<<#v<<" : ";for (auto x:v) cerr<<x<<' ';cerr<<endl;}
#define all(x) x.begin(), x.end()
#define pb push_back
#define SZ(x) ((int)(x).size())

const int inf=1000000010;
const ll INF=1000000000000001000LL;
const int mod=1000000007;
const int N=405;

int n, m, k, u, v, x, y, t, a, b, ans;
int dp[N][N][N][3], cnt[N][3];
vector<int> R, G, Y;
string S;

inline void upd(int &x, int y){ if (x>y) x=y;}

int main(){
	ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	cin>>n>>S;
	for (int i=1; i<=n; i++){
		memcpy(cnt[i], cnt[i-1], sizeof(cnt[i]));
		if (S[i-1]=='R') R.pb(i), cnt[i][0]++;
		if (S[i-1]=='G') G.pb(i), cnt[i][1]++;
		if (S[i-1]=='Y') Y.pb(i), cnt[i][2]++;
	}
	memset(dp, 63, sizeof(dp));
	dp[0][0][0][0]=dp[0][0][0][1]=dp[0][0][0][2]=0;
	for (int r=0; r<=SZ(R); r++) for (int g=0; g<=SZ(G); g++) for (int y=0; y<=SZ(Y); y++){
		if (r<SZ(R)){
			int cost=max(0, cnt[R[r]][1]-g)+max(0, cnt[R[r]][2]-y);
			upd(dp[r+1][g][y][0], min(dp[r][g][y][1], dp[r][g][y][2])+cost);
		}
		if (g<SZ(G)){
			int cost=max(0, cnt[G[g]][0]-r)+max(0, cnt[G[g]][2]-y);
			upd(dp[r][g+1][y][1], min(dp[r][g][y][0], dp[r][g][y][2])+cost);
		}
		if (y<SZ(Y)){
			int cost=max(0, cnt[Y[y]][0]-r)+max(0, cnt[Y[y]][1]-g);
			upd(dp[r][g][y+1][2], min(dp[r][g][y][0], dp[r][g][y][1])+cost);
		}
	}
	ans=inf;
	for (int i:{0, 1, 2}) upd(ans, dp[SZ(R)][SZ(G)][SZ(Y)][i]);
	if (ans==inf) ans=-1;
	cout<<ans<<"\n";
	
	return 0;
}

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