結果

問題 No.1284 Flip Game
ユーザー gazellegazelle
提出日時 2020-11-06 22:17:47
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 5,359 bytes
コンパイル時間 1,483 ms
コンパイル使用メモリ 132,148 KB
実行使用メモリ 58,240 KB
最終ジャッジ日時 2024-07-22 13:03:27
合計ジャッジ時間 4,093 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 8 ms
11,136 KB
testcase_01 WA -
testcase_02 WA -
testcase_03 AC 7 ms
11,136 KB
testcase_04 AC 7 ms
11,008 KB
testcase_05 AC 7 ms
11,136 KB
testcase_06 AC 7 ms
11,264 KB
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 WA -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
testcase_26 WA -
testcase_27 WA -
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <algorithm>
#include <cassert>
#include <random>
#include <iomanip>
#include <bitset>
#include <array>
#define FOR(i, n, m) for(ll i = (n); i < (ll)(m); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define pb push_back
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 1000000007;
constexpr long double eps = 1e-6;
 
template<typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
    os << to_string(p.first) << " " << to_string(p.second);
    return os;
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
    REP(i, v.size()) {
        if(i) os << " ";
        os << v[i];
    }
    return os;
}
 
struct modint {
    ll n;
public:
    modint(const ll n = 0) : n((n % mod + mod) % mod) {}
    static modint pow(modint a, int m) {
        modint r = 1;
        while(m > 0) {
            if(m & 1) { r *= a; }
            a = (a * a); m /= 2;
        }
        return r;
    }
    modint &operator++() { *this += 1; return *this; }
    modint &operator--() { *this -= 1; return *this; }
    modint operator++(int) { modint ret = *this; *this += 1; return ret; }
    modint operator--(int) { modint ret = *this; *this -= 1; return ret; }
    modint operator~() const { return (this -> pow(n, mod - 2)); } // inverse
    friend bool operator==(const modint& lhs, const modint& rhs) {
        return lhs.n == rhs.n;
    }
    friend bool operator<(const modint& lhs, const modint& rhs) {
        return lhs.n < rhs.n;
    }
    friend bool operator>(const modint& lhs, const modint& rhs) {
        return lhs.n > rhs.n;
    }
    friend modint &operator+=(modint& lhs, const modint& rhs) {
        lhs.n += rhs.n;
        if (lhs.n >= mod) lhs.n -= mod;
        return lhs;
    }
    friend modint &operator-=(modint& lhs, const modint& rhs) {
        lhs.n -= rhs.n;
        if (lhs.n < 0) lhs.n += mod;
        return lhs;
    }
    friend modint &operator*=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * rhs.n) % mod;
        return lhs;
    }
    friend modint &operator/=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * (~rhs).n) % mod;
        return lhs;
    }
    friend modint operator+(const modint& lhs, const modint& rhs) {
        return modint(lhs.n + rhs.n);
    }
    friend modint operator-(const modint& lhs, const modint& rhs) {
        return modint(lhs.n - rhs.n);
    }
    friend modint operator*(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * rhs.n);
    }
    friend modint operator/(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * (~rhs).n);
    }
};
istream& operator>>(istream& is, modint m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, modint m) { os << m.n; return os; }
 
#define MAX_N 1010101
long long extgcd(long long a, long long b, long long& x, long long& y) {
    long long d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}
long long mod_inverse(long long a, long long m) {
    long long x, y;
    if(extgcd(a, m, x, y) == 1) return (m + x % m) % m;
    else return -1;
}
vector<long long> fact(MAX_N+1, inf);
long long mod_fact(long long n, long long& e) {
    if(fact[0] == inf) {
        fact[0]=1;
        if(MAX_N != 0) fact[1]=1;
        for(ll i = 2; i <= MAX_N; ++i) {
            fact[i] = (fact[i-1] * i) % mod;
        }
    }
    e = 0;
    if(n == 0) return 1;
    long long res = mod_fact(n / mod, e);
    e += n / mod;
    if((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
    return (res * fact[n % mod]) % mod;
}
// return nCk
long long mod_comb(long long n, long long k) {
    if(n < 0 || k < 0 || n < k) return 0;
    long long e1, e2, e3;
    long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2), a3 = mod_fact(n - k, e3);
    if(e1 > e2 + e3) return 0;
    return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}
 
using mi = modint;
 
mi mod_pow(mi a, ll n) {
    mi ret = 1;
    mi tmp = a;
    while(n > 0) {
        if(n % 2) ret *= tmp;
        tmp = tmp * tmp;
        n /= 2;
    }
    return ret;
}
 
ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<vector<ll>> c(n, vector<ll>(n, 0));
    REP(i, n) REP(j, n) cin >> c[i][j];
    vector<vector<ll>> nc(2 * n, vector<ll>(2 * n, 0));
    REP(i, 2 * n) REP(j, 2 * n) {
        nc[i][j] = c[i % n][j % n];
    }
    
    n *= 2;

    vector<vector<ll>> dp(1 << n, vector<ll>(n, inf * inf));

    FOR(bit, 1, (1 << n)) REP(j, n) {
        bitset<18> bs(bit);
        if(bs[j] == 0) continue;
        int ppc = bs.count();
        bs[j] = 0;
        if(ppc == 1) dp[bit][j] = 0;
        else {
            REP(k, n) {
                if(abs(j - k) == n / 2) continue;
                if(bs[k] == 0) continue;
                dp[bit][j] = min(dp[bit][j], dp[bs.to_ulong()][k] + nc[k][j]);
            }
        }
    }
    ll ans = inf * inf;
    REP(j, n) ans = min(ans, dp[(1 << n) - 1][j]);
    cout << ans << endl;
    return 0;
}
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