【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

1
2
3
4
5
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1

8 0 -2

样例输出

1
15

提示

思路

代码

查看代码
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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;
}
_/_/_/_/_/ EOF _/_/_/_/_/