I. Question:
Returns the largest two-dimensional matrix (element and maximum) in a matrix. For example:
1 2 0 3 4
2 3 4 5 1
1 1 5 3 0
The largest in:
4 5
5 3
Requirements: (1) write algorithms; (2) analyze time complexity; (3) use C to write key code
Ii. Analysis:
Assume that the result of the maximum sub-matrix is the sub-matrix from row R to row K and from column I to column J,
As shown below (ARI indicates a [r] [I], assuming that the array subscript starts from 1 ):
| A11 ...... A1i ...... A1j ...... A1N |
| A21 ...... A2i ...... A2j ...... A2n |
.....
| AR1 ...... Ari ...... ARJ ...... Arn | row R
...
...... |
V
| Ak1 ...... Aki ...... Akj ...... Akn | row K...
.....
| An1 ...... Ani ...... Anj ...... Ann |
Then, we add the same columns in each row from row R to row K. We can get a one-dimensional array as follows:
(AR1 + ...... + Ak1, AR2 + ...... + Ak2 ,......, Arn + ...... + Akn)
From this we can see that the last thing we want is the maximum sub-segment and problem of this one-dimensional array,
At this point, we have converted the problem into the problem that has been solved above.
3. Source Code (the following source code is the code for finding the maximum submatrix of N rows of n columns)
# Include <iostream> <PRE name = "code" class = "html"> using namespace STD; int maxsubarray (int A [], int N) {int B = 0, sum = A [0]; for (INT I = 0; I <n; I ++) {If (B> 0) B + = A [I]; else B = A [I]; If (B> sum) sum = B;} return sum;} int maxsubmatrix (INT array [] [3], int N) {int I, J, K, max = 0, sum =-100000000; int B [3]; for (I = 0; I <n; I ++) {for (k = 0; k <n; k ++) // initialize B [] {B [k] = 0 ;}for (j = I; j <N; j ++) // Add row I to row J, and obtain the maximum value {for (k = 0; k <n; k ++) for each addition) {B [k] + = array [J] [k];} max = maxsubarray (B, k); If (max> sum) {sum = max ;}}} return sum;} int main () {int n = 3; int array [3] [3] = {1, 2, 3}, {-1,-2, -3 },{ 4, 5, 6 }}; cout <"maxsum:" <maxsubmatrix (array, n) <Endl ;}</PRE> <p>The algorithm complexity is O (n * n)