HDU 2859 Phalanx

大意：

給出一個矩陣，要求輸出最大的 延左下到右上的對角線對稱的 矩陣大小

思路：

\(now[i][j]\)代表以\((i,j)\)為左上角的長度為k矩陣是否對稱，那麼它可以由\(pre[i-1][j]\)和\(pre[i][j-1]\)轉移過來，\(pre[i][j]\)代表以\((i,j)\)為左上角的長度為k-1的矩陣是否對稱，所以如果\(pre[i-1][j]\)和\(pre[i][j-1]\)均為true，那麼只需要\(a[i][j]==a[i+k-1][j+k-1]\)即可

#include <bits/stdc++.h>

using namespace std;

const int N = 1e3 + 5;
typedef long long LL;
int n, pre[N][N], now[N][N];
char a[N][N];
int main() {
    while (cin >> n && n != 0) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cin >> a[i][j];
                pre[i][j] = 1;
            }
        }
        int res = 1;
        for (int k = 2; k <= n; k++) {
            int flag = 0;
            for (int i = 0; i <= n - k; i++) {
                for (int j = 0; j <= n - k; j++) {
                    if (a[i][j] == a[i + k - 1][j + k - 1] && pre[i + 1][j] &&
                        pre[i][j + 1])
                        now[i][j] = 1, flag = 1;
                    else
                        now[i][j] = 0;
                }
            }
            if (flag)
                res = k;
            else
                break;
            for (int i = 0; i <= n - k; i++)
                for (int j = 0; j <= n - k; j++) pre[i][j] = now[i][j];
        }
        cout << res << endl;
    }

    return 0;
}