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 00:36:21
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
TLE  
実行時間 -
コード長 4,396 bytes
コンパイル時間 2,057 ms
コンパイル使用メモリ 181,712 KB
実行使用メモリ 419,584 KB
最終ジャッジ日時 2024-09-23 06:06:20
合計ジャッジ時間 12,331 ms
ジャッジサーバーID
（参考情報） 		judge2 / 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 243 ms
410,088 KB
testcase_01 AC 241 ms
410,112 KB
testcase_02 AC 242 ms
410,112 KB
testcase_03 AC 240 ms
410,060 KB
testcase_04 AC 241 ms
409,960 KB
testcase_05 AC 240 ms
410,112 KB
testcase_06 AC 241 ms
410,024 KB
testcase_07 AC 242 ms
409,984 KB
testcase_08 TLE -
testcase_09 -- -
testcase_10 -- -
testcase_11 -- -
testcase_12 -- -
testcase_13 -- -
testcase_14 -- -
testcase_15 -- -
testcase_16 -- -
testcase_17 -- -
testcase_18 -- -
testcase_19 -- -
testcase_20 -- -
testcase_21 -- -
testcase_22 -- -
testcase_23 -- -
testcase_24 -- -
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>
#pragma GCC optimize("O3", "unroll-loops")
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;
int dp[26][2001][2001];
int H, W, K;

int count_n(int i, int j, int L)
{
    int ret = 0;
    REP(d, 26)
    {
        int tmp = dp[d][i + L][j + L] - dp[d][i][j + L] - dp[d][i + L][j] + dp[d][i][j];
        if (tmp > 0) ret++;
    }
    return ret;
}

int main()
{
    cin >> H >> W >> K;
    vector<string> C(H);
    cin >> C;
    memset(dp, 0, sizeof(dp));
    REP(i, H) REP(j, W) dp[C[i][j] - 'a'][i + 1][j + 1]++;

    REP(d, 26) REP(i, H + 1) REP(j, W)
    {
        dp[d][i][j + 1] += dp[d][i][j];
    }
    REP(d, 26) REP(i, H) REP(j, W + 1)
    {
        dp[d][i + 1][j] += dp[d][i][j];
    }

    lint ret = 0;
    REP(i, H) REP(j, W)
    {
        int dmax = min(H - i, W - j);
        int cn = count_n(i, j, dmax);
        if (cn < K) continue;

        int l = 0, r = dmax;
        while (r - l > 1)
        {
            int c = (l + r) / 2;
            if (count_n(i, j, c) < K) l = c;
            else r = c;
        }
        if (cn == K)
        {
            ret += dmax - l;
        }
        else if (count_n(i, j, r) == K)
        {
            int ll = r, rr = dmax + 1;
            while (rr - ll > 1)
            {
                int c = (ll + rr) / 2;
                if (count_n(i, j, c) > K) rr = c;
                else ll = c;

            }
            ret += ll - l;
        }
    }
    cout << ret << endl;
}
0