제출 #492838

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
492838		Carmel_Ab1	Miners (IOI07_miners)	C++17	100 / 100	100 ms	972 KiB

miners

#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//using namespace __gnu_pbds;
using namespace std;

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld, ld> pld;
typedef vector<pi> vpi;

//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T>
ostream &operator<<(ostream &os, vector<T> &a) {
    os << "[";
    for (int i = 0; i < ll(a.size()); i++) { os << a[i] << ((i != ll(a.size() - 1) ? " " : "")); }
    os << "]\n";
    return os;
}

#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << "" << p.first << " " << p.second << "";
    return os;
}

void usaco(string taskname) {
    string fin = taskname + ".in";
    string fout = taskname + ".out";
    const char *FIN = fin.c_str();
    const char *FOUT = fout.c_str();
    freopen(FIN, "r", stdin);
    freopen(FOUT, "w", stdout);
}

template<typename T>
void read(vector<T> &v) {
    int n = v.size();
    for (int i = 0; i < n; i++)
        cin >> v[i];
}

template<typename T>
vector<T> UNQ(vector<T> a) {
    vector<T> ans;
    for (T t: a)
        if (ans.empty() || t != ans.back())
            ans.push_back(t);
    return ans;
}

ll ceil(ll a, ll b) {
    ll ans = a / b;
    if (a % b)ans++;
    return ans;
}

ld log(ld a, ld b) { return log(b) / log(a); }

ll power(ll base, ll exp, ll M = 1e18) {//(base^exp)%M
    ll ans = 1;
    while (exp) {
        if (exp % 2 == 1)ans = ((ans % M) * (base % M)) % M;
        base = ((base % M) * (base % M)) % M;
        exp /= 2;
    }
    return ans;
}

string to_base(int n, int new_base) {
    string s;
    int nn = n;
    while (nn) {
        s += to_string(nn % new_base);
        nn /= new_base;
    }
    reverse(all(s));
    return s;
}

ll gcd(ll a, ll b) {
    if (a < b)swap(a, b);
    if (b == 0)return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    ll x = (a / gcd(a, b));
    return b * x;
}

vl generate_array(ll n, ll mn = 1, ll mx = 1e18 + 1) {
    if (mx == 1e18 + 1)
        mx = n;
    vl ans(n);
    for (int i = 0; i < n; i++)
        ans[i] = (mn + rand() % (mx - mn + 1));
    return ans;
}

string substr(string s, int l, int r) {
    string ans;
    for (int i = l; i <= r; i++)
        ans += s[i];
    return ans;
}


void solve();

int main() {
    GLHF;
    int t = 1;
    //cin >> t;
    while (t--)
        solve();
}

int calc(vi a){
    set<int>s={a[0],a[1],a[2]};
    s.erase(0);
    return s.size();
}
inline int get(int i,int j,int k,int l){return 64*i+16*j+4*k+l;}
void solve() {
    int n;
    cin >> n;

    string s;
    cin >> s;

    vi a(n);
    for(int i=0; i<n;i ++){
        if(s[i]=='M')
            a[i]=1;
        else if(s[i]=='B')
            a[i]=2;
        else
            a[i]=3;
    }

    int val[256];
    for(int i=0; i<256; i++)
        val[i]=0;
    for(int i=0; i<4; i++)
        for(int j=0; j<4; j++)
            for(int k=0; k<4;k++)
                val[get(i,j,k,0)]=calc({i,j,k});

    int dp[256],prv[256];

    for(int i=0; i<256; i++)
        prv[i]=-1e9;

    prv[get(a[0],0,0,0)]=1;
    prv[get(0,0,a[0],0)]=1;

    for(int i=1; i<n; i++) {
        for(int j=0; j<256; j++)
            dp[j]=-1e9;

        for (int lst11 = 0; lst11 <= 3; lst11++)
            for (int lst12 = 0; lst12 <= 3; lst12++)
                for (int lst21 = 0; lst21 <= 3; lst21++)
                    for (int lst22 = 0; lst22 <= 3; lst22++) {
                        if (!lst11 && lst12)continue;
                        if (!lst21 && lst22)continue;
                        if (a[i - 1] != lst11 && a[i - 1] != lst21)continue;

                        dp[get(a[i], lst11, lst21, lst22)] = max(
                                dp[get(a[i], lst11, lst21, lst22)],
                                prv[get(lst11, lst12, lst21, lst22)] + val[get(a[i], lst11, lst12,0)]
                        );
                        dp[get(lst11, lst12, a[i], lst21)] = max(
                                dp[get(lst11, lst12, a[i], lst21)],
                                prv[get(lst11, lst12, lst21, lst22)] + val[get(a[i], lst21, lst22,0)]
                        );
                    }
        for(int j=0; j<256; j++)
            prv[j]=dp[j];
    }
    int ans=0;

    for(int lst11=0; lst11<=3; lst11++)
        for(int lst12=0; lst12<=3; lst12++)
            for(int lst21=0; lst21<=3; lst21++)
                for(int lst22=0; lst22<=3; lst22++)
                    ans=max(ans,dp[get(lst11,lst12,lst21,lst22)]);
    out(ans)
}

컴파일 시 표준 에러 (stderr) 메시지

miners.cpp: In function 'void usaco(std::string)':
miners.cpp:57:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   57 |     freopen(FIN, "r", stdin);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~
miners.cpp:58:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   58 |     freopen(FOUT, "w", stdout);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~

• Subtask 1: 9 / 9
• Subtask 2: 9 / 9
• Subtask 3: 9 / 9
• Subtask 4: 9 / 9
• Subtask 5: 9 / 9
• Subtask 6: 7 / 7
• Subtask 7: 8 / 8
• Subtask 8: 8 / 8
• Subtask 9: 8 / 8
• Subtask 10: 8 / 8
• Subtask 11: 8 / 8
• Subtask 12: 8 / 8