「JSOI2015」symmetry
我们先考虑构造出原正方形经过 \(4\) 种轴对称变换以及 \(2\) 种旋转变换之后的正方形都构造出来，然后对所得的 \(7\) 个正方形都跑一遍二维哈希，这样我们就可以通过哈希，在 \(O(n ^ 2)\) 时间内判断原正方形中是否存在某一类型的某一大小的子正方形。
但是如果我们枚举边长，复杂度就会达到 \(O(n ^ 3)\) 级别，显然过不了。
考虑优化：我们发现对于任意一种类型的正方形，它把最外面一圈去掉之后还是满足原来的性质，所以我们可以二分来求。需要注意的是我们不好同时计算奇数边长和偶数边长的正方形，所以要二分两次。
参考代码：
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline T max(T a, T b) { return a > b ? a : b; }
template < class T > inline void read(T& s) {
s = 0; int f = 0; char c = getchar();
while ('0' > c || c > '9') f |= c == '-', c = getchar();
while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
s = f ? -s : s;
}
typedef unsigned long long ull;
const int _ = 502;
const ull base1 = 19491001, base2 = 19260817;
int n; ull pw1[_], pw2[_], h[7][_][_];
inline ull gethash(int k, int x1, int y1, int x2, int y2) {
int w = x2 - x1 + 1;
return h[k][x2][y2] - h[k][x1 - 1][y2] * pw1[w] - h[k][x2][y1 - 1] * pw2[w] + h[k][x1 - 1][y1 - 1] * pw1[w] * pw2[w];
}
inline bool check_90(int x1, int y1, int x2, int y2) {
return gethash(0, x1, y1, x2, y2) == gethash(5, y1, n - x2 + 1, y2, n - x1 + 1);
}
inline bool check_180(int x1, int y1, int x2, int y2) {
return gethash(0, x1, y1, x2, y2) == gethash(6, n - x2 + 1, n - y2 + 1, n - x1 + 1, n - y1 + 1);
}
inline bool check_diag(int x1, int y1, int x2, int y2) {
return gethash(0, x1, y1, x2, y2) == gethash(1, x1, n - y2 + 1, x2, n - y1 + 1)
|| gethash(0, x1, y1, x2, y2) == gethash(2, n - x2 + 1, y1, n - x1 + 1, y2)
|| gethash(0, x1, y1, x2, y2) == gethash(3, y1, x1, y2, x2)
|| gethash(0, x1, y1, x2, y2) == gethash(4, n - y2 + 1, n - x2 + 1, n - y1 + 1, n - x1 + 1);
}
inline bool check_4(int x1, int y1, int x2, int y2) {
return check_180(x1, y1, x2, y2) && check_diag(x1, y1, x2, y2);
}
inline bool check_8(int x1, int y1, int x2, int y2) {
return check_90(x1, y1, x2, y2) && check_diag(x1, y1, x2, y2);
}
template < class T > inline bool check(T f, int x) {
for (rg int i = x; i <= n; ++i)
for (rg int j = x; j <= n; ++j)
if (f(i - x + 1, j - x + 1, i, j)) return 1;
return 0;
}
template < class T > inline int solve(T f) {
int res = 0, l, r, mid;
l = 0, r = (n - 1) >> 1;
while (l < r) {
mid = (l + r + 1) >> 1;
if (check(f, mid << 1 | 1)) l = mid; else r = mid - 1;
}
res = max(res, l << 1 | 1);
l = 1, r = n >> 1;
while (l < r) {
mid = (l + r + 1) >> 1;
if (check(f, mid << 1)) l = mid; else r = mid - 1;
}
res = max(res, l << 1);
return res;
}
int main() {
#ifndef ONLINE_JUDGE
file("cpp");
#endif
read(n);
for (rg int i = 1; i <= n; ++i)
for (rg int j = 1; j <= n; ++j) {
scanf("%1d", &h[0][i][j]);
h[1][i][n - j + 1] = h[0][i][j];
h[2][n - i + 1][j] = h[0][i][j];
h[3][j][i] = h[0][i][j];
h[4][n - j + 1][n - i + 1] = h[0][i][j];
h[5][j][n - i + 1] = h[0][i][j];
h[6][n - i + 1][n - j + 1] = h[0][i][j];
}
pw1[0] = 1; for (rg int i = 1; i <= n; ++i) pw1[i] = pw1[i - 1] * base1;
pw2[0] = 1; for (rg int i = 1; i <= n; ++i) pw2[i] = pw2[i - 1] * base2;
for (rg int k = 0; k < 7; ++k) {
for (rg int i = 1; i <= n; ++i) for (rg int j = 1; j <= n; ++j) h[k][i][j] += h[k][i - 1][j] * base1;
for (rg int i = 1; i <= n; ++i) for (rg int j = 1; j <= n; ++j) h[k][i][j] += h[k][i][j - 1] * base2;
}
printf("%d %d %d %d %d\n", solve(check_8), solve(check_90), solve(check_4), solve(check_180), solve(check_diag));
return 0;
}