Submission #250734

#	Time	Username	Problem	Language	Result	Execution time	Memory
250734		WhaleVomit	Growing Vegetable is Fun 3 (JOI19_ho_t3)	C++14	100 / 100	196 ms	360568 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];
int dp[MAX_N][3][MAX_N][MAX_N];
 
int getcost(int c, int cnt0, int cnt1, int cnt2) { // cost to get the next c
    int idx = -1;
    if(c == 0) {
        if(cnt0 >= sz(idxOf[0])) return 1e8;
        idx = idxOf[c][cnt0];
    } else if(c == 1) {
        if(cnt1 >= sz(idxOf[1])) return 1e8;
        idx = idxOf[c][cnt1];
    } else {
        if(cnt2 >= sz(idxOf[2])) return 1e8;
        idx = idxOf[c][cnt2];
    }
    int res = idx;
    res -= min(cnt0, pref[idx][0]);
    res -= min(cnt1, pref[idx][1]);
    res -= min(cnt2, pref[idx][2]);
    return res;
}
 
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);
    M00(i, n) M00(j, n) if(i+j <= n-1) M00(k, 3) dp[n-1][k][i][j] = getcost(k, i, j, n-1-i-j);
    M00d(idx, n-1) {
        M00(c, 3) {
            for(int cnt0 = 0; cnt0 <= min(idx, pref[n-1][0]); cnt0++) {
                for(int cnt1 = 0; cnt1 <= min(idx-cnt0, pref[n-1][1]); cnt1++) {
                    int cnt2 = idx-cnt0-cnt1;
                    dp[idx][c][cnt0][cnt1] = 1e8;
                    if(cnt2 > pref[n-1][2]) continue;
                    int cost = getcost(c, cnt0, cnt1, cnt2);

                    int cnt20 = cnt0; if(c == 0) cnt20++;
                    int cnt21 = cnt1; if(c == 1) cnt21++;
                    M00(c2, 3) if(c2 != c) {
                        ckmin(dp[idx][c][cnt0][cnt1], cost + dp[idx+1][c2][cnt20][cnt21]);
                    }
                }
            }
        }
    }
    int res = 1e8;
    M00(i, 3) ckmin(res, dp[0][i][0][0]);
    if(res == 1e8) {
        finish(-1);
    }
    finish(res);
}

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