【POJ】1050.To the Max (最大子矩阵和）

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
1
2
3
4

is in the lower left corner:

9 2
-4 1
-1 8
1
2
3

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
1
2
3
4
5

Sample Output

15
1

Source

Greater New York 2001

Analysis

无长度限制的最大子段和问题是一个经典问题，只需 $O(N)$ 扫描该数列，不断把新的数加入子段，当子段和变成负数时，把当前的整个子段清空。扫描过程中出现过的最大子段和即为所求。

最大子矩阵和，就想办法将二维转成一维进行处理。将能组成矩阵的数或者和转成一维。

Code

#include 
using namespace std;

const int N = 105;
int matrix[N][N]; //二维矩阵
int temp[N]; //一维

//无长度限制的最长子段和
int maxSubSequenceSum(int *arr, int n) {
    int sum = 0;
    int ans = (int)-1e9;
    for (int i = 1; i <= n; i++) {
        if (sum + arr[i] < 0) {
            sum = 0;
        } else {
            sum += arr[i];
            ans = max(ans, sum);
        }
    }

    return ans;
}

int main() {
    int n;
    cin >> n;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> matrix[i][j];
        }
    }
    
    int ans = (int)-1e9;
    //能组成矩形的相邻数据，二维转一维
    for (int i = 1; i <= n; i++) { //每行选择的数的开始位置
        for (int j = i; j <= n; j++) { //每行选择的数的终止位置
            for (int k = 1; k <= n; k++) { //枚举行（转一维）
                temp[k] = 0;
                for (int m = i; m <= j; m++) {
                    temp[k] += matrix[k][m];
                }
            }
            ans = max(ans, maxSubSequenceSum(temp, n));
        }
    }

    cout << ans << endl;
    return 0;
}

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