HDU 4618 Palindrome Sub-Array

Palindrome Sub-Array

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 503    Accepted Submission(s): 256

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

  

   1
5 10
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 3 2 4 5 6 7 8
1 2 3 9 10 4 5 6 7 8
  

 

Sample Output

  

   4
  

 

Source

2013 Multi-University Training Contest 2

 

Recommend

zhuyuanchen520

 

题意： 有一个n * m的矩阵， 问最大边长的回文矩阵。

回文矩阵是 每一行都是回文串， 每一列都是回文串。

思路：乱搞

#include <cstdio>  
#include <algorithm>  
using namespace std;  
  
//回文矩阵特点 水平对称，垂直对称， 离矩阵中心距离相等的值必须相等。  
//然后乱搞  
const int v = 300 + 5;  
int t, n, m, num[v][v], ans;  
bool is_ok(int ii, int jj, int len) {  
    int r_len = len;  
    for(int i = ii; i < ii + len / 2; ++i) {  
        int c_len = len;  
        for(int j = jj; j < jj + len / 2; ++j) {  
            if(num[i][j] == num[i][j + c_len - 1] && num[i][j] == num[i + r_len - 1][j] && num[i][j] == num[i + r_len - 1][j + c_len - 1])  
                c_len -= 2;  
            else  
                return false;  
        }  
        r_len -= 2;  
    }  
    return true;  
}  
int main() {  
    int i, j;  
    scanf("%d", &t);  
    while(t--) {  
        ans = 1;  
        scanf("%d%d", &n, &m);  
        for(i = 0; i < n; ++i)  
            for(j = 0; j < m; ++j)  
                scanf("%d", &num[i][j]);  
        for(i = 0; i < n; ++i)  
            for(j = 0; j < m; ++j)  
                for(int k = min(n, m); k > ans; --k) {  
                    if(i + k > n || j + k > m)  
                        continue;  
                    if(is_ok(i, j, k)) {  
                        ans = max(ans, k);  
                        break;  
                    }  
                }  
        printf("%d\n", ans);  
    }  
}