【题解】POJ-1050 To the Max

To the Max (POJ - 1050)

题面

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.

输入

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].

输出

Output the sum of the maximal sub-rectangle.

样例输入

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

8  0 -2

样例输出

15

提示

无

思路

代码

using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 1e2 + 5;

int a[N][N] = {{0}};
int f[N][N][N] = {{{0}}};
int n;

int main(void) {
    scanf("%d", &n);
    for (int i=0; i<n; i++) {
        for (int j=0; j<n; j++)
            scanf("%d", &a[i][j]);
    }
    int mx = -inf;
    for(int i=0; i<n; i++) {
        for(int j=0; j<n; j++) {
            int sum = 0;
            for(int k=j; k<n; k++) {
                sum += a[i][k];
                f[i][j][k] = max(f[i-1][j][k]+sum, sum);
                mx = max(mx, f[i][j][k]);
            }
        }
    }
    printf("%d\n", mx);
    return 0;
}