Submission #1232075

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Source Code Expand

#include <cstdio>
#include <cstring>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <string>
#include <queue>
#include <map>
#include <stack>
#include <vector>
#include <algorithm>
#include <set>
#include <deque>
#include <utility>
#include <chrono>
#include <sstream>
#include <iomanip>
using namespace std;
#define MOD 1000000007
#define PI 3.14159265358979
#define rep(i, n) for (int (i) = 0; (i) < (int)(n); (i)++)
#define rer(i, l, r) for (int (i) = (int)(l); (i) <= (int)(r); (i)++)
#define reu(i, l, r) for (int (i) = (int)(l); (i) < (int)(r); (i)++)
#define each(i, v) for (auto i : v)
#define D(x) cout << x << endl
#define d(x) cout << x
#define all(x) (x).begin(), (x).end()
#define pub(x) push_back(x)
#define pob() pop_back()
#define puf(x) push_front(x)
#define pof() pop_front()
#define mp(x, y) make_pair((x), (y))
#define fi first
#define se second
#define setp(x) setprecision(x)
#define mset(m, v) memset(m, v, sizeof(m))
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
typedef long long ll;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<long long> vll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef pair<int, int> pii;
typedef pair<long, long> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
template<typename T, typename U> inline void amin(T &x, U y) { if (y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if (x < y) x = y; }

//static const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
//#define int ll

int main() { 
        int R, C, K;
        cin >> R >> C >> K;
        vs s(R);
        rep(i, R) cin >> s[i];
        vector<vi> sum(R, vi(C + 1, 0));
        rep(i, R) rep(j, C) {
                sum[i][j + 1] = sum[i][j] + (s[i][j] == 'x');
        }
        int ans = 0;
        rer(y, K - 1, R - K) rer(x, K - 1, C - K) {
                bool f = true;
                rer(dy, -K + 1, K - 1) {
                        int yy = y + dy, t = K - abs(dy);
                        f &= sum[yy][x + t] - sum[yy][x - t + 1] == 0;
                }
                ans += f;
        }
        D(ans);
        return 0;
}

Submission Info

Judge Result