結果

問題 No.866 レベルKの正方形
ユーザー hitonanodehitonanode
提出日時 2019-08-17 00:53:01
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
MLE  
実行時間 -
コード長 3,664 bytes
コンパイル時間 1,786 ms
コンパイル使用メモリ 173,332 KB
実行使用メモリ 597,892 KB
最終ジャッジ日時 2024-09-24 07:22:28
合計ジャッジ時間 33,665 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
このコードへのチャレンジ
（要ログイン）

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 37 ms
97,248 KB
testcase_01 AC 36 ms
97,184 KB
testcase_02 AC 36 ms
97,352 KB
testcase_03 AC 37 ms
97,184 KB
testcase_04 AC 37 ms
97,200 KB
testcase_05 AC 38 ms
97,240 KB
testcase_06 AC 37 ms
97,168 KB
testcase_07 AC 37 ms
97,236 KB
testcase_08 MLE -
testcase_09 MLE -
testcase_10 MLE -
testcase_11 MLE -
testcase_12 MLE -
testcase_13 MLE -
testcase_14 MLE -
testcase_15 MLE -
testcase_16 MLE -
testcase_17 MLE -
testcase_18 MLE -
testcase_19 MLE -
testcase_20 MLE -
testcase_21 MLE -
testcase_22 AC 37 ms
97,276 KB
testcase_23 AC 36 ms
97,168 KB
testcase_24 AC 36 ms
97,200 KB
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;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#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> 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); }
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
///// This part below is only for debug, not used /////
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; }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
///// END /////


using lint = long long int;
vector<int> dp[2001][2001];
int H, W, K;

constexpr int INF = 100000;
int main()
{
    cin >> H >> W >> K;
    REP(i, H + 1) REP(j, W + 1) dp[i][j].assign(27, INF);
    REP(i, H) REP(j, W)
    {
        char c;
        cin >> c;
        dp[i][j][c - 'a'] = 1;
    }
    IREP(i, H) IREP(j, W) REP(d, 26)
    {
        mmin(dp[i][j][d], min({dp[i + 1][j][d], dp[i][j + 1][d], dp[i + 1][j + 1][d]}) + 1);
    }
    lint ret = 0;
    REP(i, H) REP(j, W)
    {
        sort(dp[i][j].begin(), dp[i][j].end());
        int d_max = min(H - i, W - j);
        ret += min(dp[i][j][K] - 1, d_max) - min(dp[i][j][K - 1] - 1, d_max);
    }
    cout << ret << endl;
}
0