POJ - 1050 To the Max (DP + Max Matrix Sum/Subsegment Sum)

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 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.

Input

The input consists of an N * 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-dimensional 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].

Output

Output the sum of the maximal sub-rectangle.

Sample Input

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

8 0 -2

Sample Output

15

Problem solving ideas:

The two-dimensionalization of the maximum subsection and the problem
Basic idea: dimensionality reduction processing
Implementation: First select the row where the matrix is ​​located (using a double loop to consider all selection situations)
　　　After selecting the row, it is the problem of selecting which columns Now, take the selected row 2 ~ 4 as an example. That is, i=2 j=4, the maximum value can be obtained by finding the combination of these columns. Since it is always 2 3 4 rows, we can "bundle" these 3 rows to find 4 (9-4-1) ,11(8+2+1),-10(-6-4+0), 7(7+2-2) The maximum sub-segment sum, this problem has been successfully transformed into a one-dimensional case!
　　　After this step, the problem has been successfully transformed into the largest subsection and problem.

Code:

//最大子段和问题的二维化
/*降维*/
#include <cstdio>
#include <cstring>
using namespace std;

int map[110][110];
int temp[110];
int main(){
    int n;
    int i , j ,max;
    while(scanf("%d" ,&n) != EOF){
        memset(map , 0 , sizeof(map));

        for( i = 1 ; i <= n ; i ++){
            for(j = 1 ; j <= n ; j ++){
                scanf("%d" , &map[i][j]);
                map[i][j] = map[i][j] + map[i-1][j];    //这一步实际上已经实现了一个降维的过程
            }
        }

        max = map[1][1];
        for(i = 1 ; i <= n ; i ++){    //i 表示起始行 ， j 表示结束行
            /*行数确定的情况下，只要确定一下列数就可以 ， 这时候就是将 选定行中的同一列中的数字相加
                便能够得到一个一维的数组 ， 到这步位置就已经将这个问题降维 成 最大子段和 问题了。*/
            for(j = i ; j <= n ; j ++){
                memset(temp , 0 , sizeof(temp));
                for(int k = 1 ; k <= n ; k++)         //temp 即为降维之后的一维数组
                    temp[k] = map[j][k] - map[i-1][k];

                for(int k = 1 ; k  <= n ; k ++ ){
                    if(temp[k-1] >= 0)
                        temp[k] = temp[k -1] + temp[k];
                    if(temp[k] > max)
                        max = temp[k];
                }
            }
        }
        printf("%d\n",max);
    }
    return 0;
}