結果

問題 No.819 Enjapma game
ユーザー hitonanodehitonanode
提出日時 2019-04-19 22:12:23
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,944 bytes
コンパイル時間 1,843 ms
コンパイル使用メモリ 176,904 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-09-22 20:28:41
合計ジャッジ時間 2,592 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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 1 ms
6,812 KB
testcase_01 AC 2 ms
6,944 KB
testcase_02 AC 2 ms
6,944 KB
testcase_03 WA -
testcase_04 WA -
testcase_05 AC 2 ms
6,940 KB
testcase_06 AC 2 ms
6,940 KB
testcase_07 AC 2 ms
6,940 KB
testcase_08 WA -
testcase_09 AC 2 ms
6,940 KB
testcase_10 WA -
testcase_11 AC 4 ms
6,944 KB
testcase_12 WA -
testcase_13 WA -
testcase_14 AC 3 ms
6,944 KB
testcase_15 WA -
testcase_16 AC 2 ms
6,940 KB
testcase_17 WA -
testcase_18 WA -
testcase_19 AC 2 ms
6,944 KB
testcase_20 WA -
testcase_21 WA -
testcase_22 AC 4 ms
6,940 KB
testcase_23 WA -
testcase_24 AC 2 ms
6,940 KB
testcase_25 WA -
testcase_26 WA -
testcase_27 AC 2 ms
6,944 KB
testcase_28 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 <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#define FI first
#define SE second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))

int H, W;

int count_o(vector<string> S)
{
    int ret = 0;
    for (auto vec : S) for (auto c : vec) ret += (c == 'o');
    return ret;
}

int main()
{
    cin >> H >> W;
    vector<string> S(H);
    REP(h, H) cin >> S[h];

    if (count_o(S) % 2)
    {
        // o が奇数 先手必勝
        puts("NO");
    }
    else
    {
        // o が偶数　残り移動回数が奇数なら先手必勝
        lint ret = 0;
        reverse(ALL(S));
        vector<int> mv;
        REP(h, H) REP(w, W)
        {
            if (S[h][w] == 'o') mv.push_back(h + w);
        }

        priority_queue<int, vector<int>, greater<int>> pq;
        REP(h, H) REP(w, W) pq.push(h + w);

        ret = accumulate(ALL(mv), 0);
        REP(_, mv.size()) ret -= pq.top(), pq.pop();
        if (ret % 2)
        {
            puts("NO");
        }
        else
        {
            puts("YES");
        }
        }
    
}
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