#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using ll = long long; using Pll = pair; using Pii = pair; constexpr ll MOD = 1000000007; constexpr long double EPS = 1e-10; constexpr int dyx[4][2] = { { 0, 1}, {-1, 0}, {0,-1}, {1, 0} }; int counts[2001][2001][26]; int count_level(int i, int j, int r) { int level = 0; for(int k=0;k<26;++k) { if(counts[i+r-1][j+r-1][k] - counts[i+r-1][j-1][k] - counts[i-1][j+r-1][k] + counts[i-1][j-1][k] > 0) { ++level; } } return level; } int main() { std::ios::sync_with_stdio(false); cin.tie(0); int h,w,k; cin >> h >> w >> k; vector> a(h+1, vector(w+1, -1)); for(int i=1;i<=h;++i) { string s; cin >> s; for(int j=1;j<=w;++j) { a[i][j] = int(s[j-1] - 'a'); } } int kr = 1; // k を達成するのに最低限必要な半径 while(kr*kr < k) ++kr; fill_n(counts[0][0], (h+1)*(w+1)*26, 0); for(int i=1;i<=h;++i) { for(int j=1;j<=w;++j) { for(int k=0;k<26;++k) { counts[i][j][k] = counts[i-1][j][k] + counts[i][j-1][k] - counts[i-1][j-1][k]; } ++counts[i][j][a[i][j]]; } } int ans = 0; for(int i=1;i<=h;++i) { for(int j=1;j<=w;++j) { int ok = 0, ng = min(h, w); // ok: 半径 ok のとき level <= k; // ng: 半径 ng のとき level > k; while(ng-ok > 1) { int r = (ok+ng) / 2; if(i+r-1 > h || j+r-1 > w) { ng = r; continue; } int level = count_level(i, j, r); if(level > k) { ng = r; } else { ok = r; } } int max_r_k = ok; ng = min(h, w); ok = 0; // ok: 半径 ok のとき level <= k-1; // ng: 半径 ng のとき level > k-1; while(ng-ok > 1) { int r = (ok+ng) / 2; if(i+r-1 > h || j+r-1 > w) { ng = r; continue; } int level = count_level(i, j, r); if(level > k-1) { ng = r; } else { ok = r; } } int max_r_k1 = ok; ans += max(0, max_r_k-max_r_k1); // cerr << i << "," << j << ": " << max_r_k << " " << max_r_k1 << endl; } } cout << ans << endl; }