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.
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
解题思路:
最大子矩阵,首先一行数列很简单求最大的子和,我们要把矩阵转化成一行数列,
就是从上向下在输入的时候取和,a[i][j]表示在J列从上向下的数和,这样就把一列转化成了一个点,
再用双重,循环,任意i行j列开始的一排数的最大和,就是最终的最大和
注意:如果m<0,则就不需要继续加了
程序代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <cstring> 3 using namespace std; 4 int a[110][110]; 5 int main() 6 { 7 int n,c; 8 while( scanf("%d",&n)==1) 9 { 10 memset(a,0,sizeof(a)); 11 for(int i=1;i<=n;i++) 12 for(int j=1;j<=n;j++) 13 { 14 scanf("%d",&c); 15 a[i][j]=a[i-1][j]+c; 16 } 17 int sum=0; 18 for(int i=1;i<=n;i++) 19 for(int j=i;j<=n;j++) 20 { 21 int m=0; 22 for(int k=1;k<=n;k++) 23 { 24 int t=a[j][k]-a[i-1][k]; 25 m+=t; 26 if(m<0) m=0; 27 if(sum<m) sum=m; 28 } 29 } 30 31 printf("%d ",sum); 32 } 33 return 0; 34 }