Submission #250720

#	Time	Username	Problem	Language	Result	Execution time	Memory
250720		WhaleVomit	Growing Vegetable is Fun 3 (JOI19_ho_t3)	C++14	75 / 100	1096 ms	61192 KiB

joi2019_ho_t3

#include <bits/stdc++.h>
using namespace std;

#define f first
#define s second
#define pb push_back
#define mp make_pair
#define all(v) v.begin(), v.end()
#define sz(v) (int)v.size()

#define MOO(i, a, b) for(int i=a; i<b; i++)
#define M00(i, a) for(int i=0; i<a; i++)
#define MOOd(i,a,b) for(int i = (b)-1; i >= a; i--)
#define M00d(i,a) for(int i = (a)-1; i>=0; i--)

#define FAST ios::sync_with_stdio(0); cin.tie(0);
#define finish(x) return cout << x << '\n', 0;
#define dbg(x) cerr << ">>> " << #x << " = " << x << "\n";
#define _ << " _ " <<

template<class T> bool ckmin(T& a, const T& b) {return b < a ? a = b, 1 : 0;}
template<class T> bool ckmax(T& a, const T& b) {return a < b ? a = b, 1 : 0;}

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef pair<ld,ld> pd;
typedef complex<ld> cd;

const int MAX_N = 404;
int n;
int arr[MAX_N];
int pref[MAX_N][3];
vi idxOf[3];
map<pair<pi, vi>, int> dp;

int getcost(int c, vi cnt) { // cost to get the next c
    if(cnt[c] >= sz(idxOf[c])) return 1e8; // impossible
    int idx = idxOf[c][cnt[c]];
    int res = idx;
    M00(k, 3) res -= min(cnt[k], pref[idx][k]);
    return res;
}

int getdp(int idx, int c, vi cnt) {
    pair<pi, vi> tmp = mp(mp(idx, c), cnt);
    if(dp.count(tmp)) return dp[tmp];
    int cost = getcost(c, cnt);
    if(cost == 1e8) {
        dp[tmp] = 1e8;
        return dp[tmp];
    }
    if(idx == n-1) {
        dp[tmp] = cost;
        return dp[tmp];
    }
    cnt[c]++;
    int res = 1e8;
    M00(c2, 3) if(c2 != c) {
        ckmin(res, cost + getdp(idx+1, c2, cnt));
    }
    dp[tmp] = res;
    return dp[tmp];
}

int main() { FAST
    string s;
    cin >> n >> s;
    M00(i, n) {
        if(s[i] == 'R') arr[i] = 0;
        else if(s[i] == 'Y') arr[i] = 1;
        else arr[i] = 2;
    }
    M00(c, 3) {
        pref[0][c] = (arr[0] == c);
        MOO(i, 1, n) pref[i][c] = pref[i-1][c] + (arr[i] == c);
    }
    M00(i, n) idxOf[arr[i]].pb(i);
    int res = 1e8;
    vi v(3,0);
    M00(i, 3) ckmin(res, getdp(0, i, v));
    if(res == 1e8) {
        finish(-1);
    }
    finish(res);
}

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