[ZJOI2009]对称的正方形 (二维哈希+二分)

题意

求矩阵中上下对称且左右对称的正方形子矩阵的个数。

思路

快速对比矩阵是否上下对称或者左右对称可以考虑对二维矩阵哈希。

哈希之后通过枚举正方形中心点位置，对正方形边长进行二分。

在枚举正方形中心点时，需考虑正方形边长分别为奇偶的情况，即中心点为格子交接点还是格子中心点。

二维矩阵哈希值维护：

(hash[x][y] = hash[x][y-1] imes base1 + hash[x - 1][y] imes base2 + value[x][y])

二维矩阵哈希值求解：

(ans = hash[x][y] - hash[x - len_x][y] imes fac1[len_x] - hash[x][y -len_y] imes fac2[len_y] + hash[x - len_x][y - len_y] imes fac1[len_x] imes fac2[len_y])

其中 (fac1), (fac2) 分别为 (base1), (base2) 的 (len) 次方

代码

#include <cstdio>
#include <algorithm>
#include <iostream>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
const int maxn = 2010;
const int base1 = 87;
const int base2 = 31;

int n, m;
int mp[maxn][maxn];
int lr[maxn][maxn];
int up[maxn][maxn];

ull hash_mp[maxn][maxn];
ull hash_lr[maxn][maxn];
ull hash_up[maxn][maxn];
ull fac1[maxn], fac2[maxn];

bool check(int x, int y, int len) {
    if (x > n || y > m) return 0;
    if (x < len || y < len) return 0;
    ull ans1 = hash_mp[x][y] - hash_mp[x - len][y] * fac2[len] - hash_mp[x][y - len] * fac1[len]
            + hash_mp[x - len][y - len] * fac1[len] * fac2[len];
    int cow_y = m - (y - len);
    ull ans2 = hash_lr[x][cow_y] - hash_lr[x - len][cow_y] * fac2[len] - hash_lr[x][cow_y - len] * fac1[len]
            + hash_lr[x - len][cow_y - len] * fac1[len] * fac2[len];
    if (ans1 != ans2) return 0;
    int row_x = n - (x - len);
    ull ans3 = hash_up[row_x][y] - hash_up[row_x - len][y] * fac2[len] - hash_up[row_x][y - len] * fac1[len]
            + hash_up[row_x - len][y - len] * fac1[len] * fac2[len];
    if (ans1 != ans3) return 0;
    return (ans1 == ans3 && ans2 == ans3);
}

void solve() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j)
            scanf("%d", &mp[i][j]);
    }

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            lr[i][j] = mp[i][m - j + 1];
            up[i][j] = mp[n - i + 1][j];
        }
    }

    fac1[0] = fac2[0] = 1;
    for (int i = 1; i <= n; ++i) fac1[i] = fac1[i - 1] * base1;
    for (int i = 1; i <= m; ++i) fac2[i] = fac2[i - 1] * base2;

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
           hash_mp[i][j] = hash_mp[i][j - 1] * base1 + mp[i][j];
           hash_lr[i][j] = hash_lr[i][j - 1] * base1 + lr[i][j];
           hash_up[i][j] = hash_up[i][j - 1] * base1 + up[i][j];
        }
    }

    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            hash_mp[i][j] += hash_mp[i - 1][j] * base2;
            hash_lr[i][j] += hash_lr[i - 1][j] * base2;
            hash_up[i][j] += hash_up[i - 1][j] * base2;
        }
    }

    ll ans = 0;
    int R = max(n, m);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            int l = 1, r = R;
            while (l <= r) {
                int mid = l + r >> 1;
                if (check(i + mid, j + mid, mid << 1 | 1)) {
                    l = mid + 1;
                } else {
                    r = mid - 1;
                }
            }
//            cout << "~ " << l << " " << r << endl;
            ans += r;
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            int l = 1, r = R;
            while (l <= r) {
                int mid = l + r >> 1;
                if (check(i + mid, j + mid, mid << 1)) {
                    l = mid + 1;
                } else {
                    r = mid - 1;
                }
            }
//            cout << "~~ " << i << " " << j << " " << l << " " << r << endl;
            ans += r;
        }
    }
    printf("%lld
", ans + n * m);
}

int main() {
    int T = 1;
//    scanf("%d", &T);
    while (T--)
        solve();
    return 0;
}