Openjudge 2766 maximum submatrix

1. Link:

Http://bailian.openjudge.cn/practice/2766

2. Question:

Total time limit:

1000 ms

Memory limit:

65536kb

Description

The size of the known matrix is defined as the sum of all elements in the matrix. Given a matrix, your task is to find the largest non-null (size at least 1*1) submatrix.

For example, the following 4*4 Matrix

0-2-7 0
9 2-6 2
-4 1-4 1
-1 8 0-2

Is

9 2
-4 1
-1 8

The size of this Sub-matrix is 15.

Input

The input is a matrix of N * n. The first line of the input is n (0 <n <= 100 ). The following rows are in sequence (first, N Integers of the first row are given from left to right, and then n Integers of the second row are given from left to right ......) The N2 integers in the matrix are given. integers are separated by spaces (spaces or empty rows ). The range of integers in the known matrix is [-127,127].

Output

The maximum size of the output sub-matrix.

Sample Input

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample output

15

Source

Translated from greater New York 2001

3. Ideas:

The largest expanded field and. Traverse all possible rows first K = 1-N. Then calculate the sum of each column in the K-row matrix and convert it to one-dimensional. The maximum value is obtained using the maximum field sum method.

4. Code:

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4  5 using namespace std; 6  7 int main() 8 { 9     //freopen("C://input.txt","r",stdin);10 11     int i,j,k;12 13     int n;14     cin >> n;15 16     int **arr_matrix = new int*[n];17     for(i = 0; i < n; ++i) arr_matrix[i] = new int[n];18 19 20     for(i = 0;i < n; ++i)21     {22         for(j = 0;j < n; ++j)23         {24             cin >> arr_matrix[i][j];25         }26     }27 28     int *arr_temp = new int[n];29 30     int *dp = new int[n];31 32     int max_sum = arr_matrix[0][0];33     for(k = 0; k < n; ++k)34     {35         memset(arr_temp,0,sizeof(int) * n);36         for(j = 0; j < n; ++j)37         {38             for(i = 0; i < k; ++i) arr_temp[j] += arr_matrix[i][j];39         }40 41         for(i = k; i < n; ++i)42         {43             for(j = 0; j < n; ++j) arr_temp[j] += arr_matrix[i][j];44 45             memset(dp,0,sizeof(int) * n);46             dp[0] = arr_temp[0];47             for(j = 1; j < n; ++j)48             {49                 dp[j] = ((dp[j - 1] + arr_temp[j]) > arr_temp[j]) ? (dp[j - 1] + arr_temp[j]) : arr_temp[j];50                 if(max_sum < dp[j]) max_sum = dp[j];51             }52 53             for(j = 0; j < n; ++j) arr_temp[j] -= arr_matrix[i - k][j];54         }55     }56 57     cout << max_sum << endl;58 59     delete [] dp;60 61     delete [] arr_temp;62 63     for(i = 0; i < n; ++i) delete [] arr_matrix[i];64     delete [] arr_matrix;65 66     return 0;67 }