F - To the Max

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

 1 #include<cstdio>

 2 #include<string.h>

 3 using namespace std;

 4 int map[120][120];

 5 int main()

 6 {

 7     int n;

 8     int flag;

 9     int sum;

10     while(scanf("%d",&n)!=EOF)

11     {

12         memset(map,0,sizeof(map));

13         for(int i=1; i<=n; i++)

14             for(int j=1; j<=n; j++)

15             {

16                 scanf("%d",&flag);

17                 map[i][j]+=map[i][j-1]+flag;//先计算每一行的前J列的和

18         for(int i=1; i<=n; i++)

19             for(int j=1; j<=i; j++)//确定子矩阵的左右边界

20             {

21                 int sum=0;

22                 for(int k=1; k<=n; k++)//行数

23                 {

24                     if(sum<0)

25                         sum=0;//如果前几行的和是小于0的，就令sum=0;因为负数越加越小。

26                     sum+=(map[k][i]-map[k][j-1]);//一行一行地加起来，第k行的[i,j]列

27                     if(sum>max)

28                         max=sum;//找出最大值！！！

29                 }

30                             

31             }

32             printf("%d\n",max);

33     }

34     return 0;

35 }