HDU 4618 palindrome sub-array (maximum palindrome matrix)

Palindrome sub-array

Time Limit: 3000/1000 MS (Java/others) memory limit: 65535/32768 K (Java/Others)
Total submission (s): 162 accepted submission (s): 72


Problem description a palindrome sequence is a sequence which is as same as its reversed order. for example, 1 2 3 2 1 is a palindrome sequence, but 1 2 3 2 2 is not. given a 2-D array of N rows and M columns, your task is to find a maximum sub-array of P rows and P columns,
Of which each row and each column is a palindrome sequence.

Input the first line of input contains only one integer, T, the number of test cases. Following t blocks, each block describe one test case.
There is two integers n, m (1 <= n, m <= 300) separated by one white space in the first line of each block, representing the size of the 2-D array.
Then n lines follow, each line contains M integers separated by white spaces, representing the elements of the 2-D array. all the elements in the 2-D array will be larger than 0 and no more than 31415926.

Output for each test case, output P only, the size of the maximum sub-array that you need to find.

Sample Input

15 101 2 3 3 2 4 5 6 7 81 2 3 3 3 2 5 6 7 81 2 3 3 2 4 5 6 7 81 2 3 3 2 4 5 6 7 81 2 3 9 10 4 5 6 7 8
 

Sample output

 
4
 

Source2013 multi-university training Contest 2

Recommendzhuyuanchen520 direct violence, but pay attention to the break to prevent timeout! No technical skills!

# Include <iostream> # include <string. h> # include <stdio. h ># include <algorithm> using namespace STD; int map [330] [330]; int rec [330]; int n, m; int ans; int is_ OK (int I, int J, int Len) {int K = 0; Len --; while (k <Len) {If (Map [I] [J + k]! = Map [I] [J + Len]) return 0; k ++, Len --;} return 1;} int is_ OK _ver (int I, Int J, int Len) {int K = I + len-1; while (I <k) {If (Map [I] [J]! = Map [k] [J]) return 0; I ++, k --;} return 1;} int is_matrix (int I, Int J, int Len) {int K, r; For (k = 0; k <Len; k ++) {If (! Is_ OK (I + k, J, Len) return 0; If (! Is_ OK _ver (I, j + k, Len) return 0;} return 1;} int find_ans (int I, Int J) {int P = n-I, Q = m-J, Len; Len = P <q? P: Q; If (ANS> = Len) return 0; while (LEN> ans) {If (is_matrix (I, j, Len) {ans = Len; return 0;} Len --;} return 1;} int main () {int T, I, J, K; scanf ("% d", & T ); while (t --) {ans = 1; scanf ("% d", & N, & M); for (I = 0; I <n; I ++) for (j = 0; j <m; j ++) scanf ("% d", & map [I] [J]); for (I = 0; I <n; I ++) for (j = 0; j <m; j ++) {find_ans (I, j) ;} printf ("% d \ n ", ans);} return 0 ;}