The sum of the maximum child matrices of HDU 1081 and the DP

To the max Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 8210 accepted submission (s): 3991


Problem descriptiongiven a two-dimen=array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1x1 or greater located within the whole array. the sum of a rectangle is the sum of all the elements in that rectangle. in this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

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

Is in the lower left corner:

9 2
-4 1
-1 8

And has a sum of 15.
 
Inputthe input consists of an n x n Array of integers. the input begins with a single positive integer n on a line by itself, indicating the size of the square two-dimen=array. this is followed by N 2 integers separated by whitespace (spaces and newlines ). these are the N 2 integers of the array, presented in row-Major Order. that is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. the numbers in the array will be in the range [-127,127].
 
Outputoutput the sum of the maximal sub-rectangle.
 
Sample Input

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

 
Sample output

15 The question is clear: give an n * n matrix and find its sub-matrix (element and maximum); (1) first, we need to know how to calculate the maximum child segment and of a one-dimensional array;
Int sum = 0, max =-9999999; For (INT I = 1; I <= N; I ++) // n elements are used in the array, after traversing it, You can {If (sum <0) sum = 0; // If sum is smaller than 0, you can skip the previous steps and start from scratch, initialize 0 sum + = A [I]; If (sum> MAX) max = sum; // update Max continuously. The final Max is the maximum field and}
(2) understand the maximum field and solution of one dimension, and then look at the two-dimensional. The classic method is to compress the two-dimensional data into one-dimensional data. If this compression process is clear, it will be easy; for example, the matrix of 4*4 is 0-2-7 0 9 2-6 2-4 1-4 1-1 8 0-2 1. regard the first column as one-dimensional to find the maximum Max; 2. merge the first column and the second column as one-dimensional Max 3. merge the first, second, and third columns as one-dimensional maxversions. 4. merge the first, second, third, and fourth columns as one-dimensional maxversions;
5. Think of the second column as the one-dimensional maximum value 6. merge the second column and the third column into one-dimensional maximum value 7. merge the second column, third column, and fourth column into one-dimensional Maximum Value
8. Think of the third column as one-dimensional, evaluate the maximum value 9. Merge the third column and the fourth column to obtain the maximum value;
10. Consider the fourth column as one-dimensional, and find the final value ~~~~~~ The compression process is like this ~~ It should be understandable !!! Before compression, perform the following operations: DP [I] [J] indicates the sum of elements in column J before row I. For example, when merging the second and third columns, the following operations are performed: DP [1] [3]-DP [1] [2-1] indicates the sum of the first three elements of the first row and the sum of the first element of the first row. one-dimensional Processing ~~~ And so on
# Include <iostream> # include <stdio. h> # include <string. h ># include <string> const int maxn = 110; using namespace STD; int DP [maxn] [maxn]; int main () {int N; while (CIN> N) {memset (DP, 0, sizeof (DP); For (INT I = 1; I <= N; I ++) // DP [I] [J] indicates the first J elements of row I for (Int J = 1; j <= N; j ++) {CIN> DP [I] [J]; DP [I] [J] + = DP [I] [J-1];} int max =-99999999; for (INT I = 1; I <= N; I ++) for (Int J = I; j <= N; j ++) {int sum = 0; for (int K = 1; k <= N; k ++) {If (sum <0) sum = 0; sum + = DP [k] [J]-DP [k] [I-1]; // DP [k] [J]-DP [k] [I-1] indicates and if (sum> MAX) of elements from column I to column J of row K) max = sum ;}} cout <max <Endl ;}return 0 ;}