最大连续和 Medium

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

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
const int N = 100 + 5;
int mat[N][N];

int maxsum(int n){
    int matsum[N][N], ret, sum;
    int i, j, k;
    for(i = 0; i < n; i++)
        for(matsum[i][j = 0] = 0; j < n; j++) 
            matsum[i][j + 1] = mat[i][j] + matsum[i][j];
    for(ret = mat[0][j = 0]; j < n; j++)
        for(k = j; k < n; k++)
            for(sum = 0, i = 0; i < n; i++)
            sum = (sum > 0? sum:0) + matsum[i][k + 1] - matsum[i][j], ret = (sum > ret?sum:ret);
    return ret;
}

int main(){
    int n;
    while(scanf("%d", &n) == 1){
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++) scanf("%d", &mat[i][j]);
        printf("%d\n", maxsum(n));
    }
}