제출 #229252

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
229252		534351	HicCup (FXCUP4_hiccup)	C++17	100 / 100	74 ms	42872 KiB

hiccup

#include "hiccup.h"
#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

const int MAXN = 1000013;

int N, M, K;
pii pos[MAXN];
int val[MAXN];
int parent[MAXN];
int freq[MAXN];
vi edge[MAXN];
int ans = MAXN;

void proc(int u)
{
    int c = 0, n = 0;
    for (int v : edge[u])
    {
        c += freq[v];
        n++;
        ckmin(ans, c / n);
    }
}
int HicCup(string str)
{
    N = SZ(str);
	if (str[0] == '!') return -1;
    vi stk; stk.PB(N);
    FORD(i, N, 0)
    {
        char c = str[i];
        if (c == '!')
        {
            if (str[i - 1] == 'H')
            {
                return -1;
            }
            K++;
        }
        else if (c == 'H')
        {
            if (stk.empty()) return -1;
            val[i] = stk.back();
            stk.pop_back();
            if (stk.empty()) return -1;
            parent[val[i]] = stk.back();
            edge[parent[val[i]]].PB(val[i]);
            // cerr << "edge " << parent[val[i]] << " - > " << val[i] << endl;
            K = 0;
        }
        else if (c == 'C')
        {
            val[i] = M;
            M++;
            stk.PB(val[i]);
            freq[val[i]] = K;
            // cerr << val[i] << " has " << K << endl;
            K = 0;
        }
        else return -1;
    }
    if (SZ(stk) != 1 || stk[0] != N)
    {
        return -1;
    }
    FOR(i, 0, M)
    {
        proc(i);
    }
    proc(N);
    return ans;
}

• Subtask 1: 24 / 24
• Subtask 2: 76 / 76