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F - To the Max

2024-07-15 21:00:37   224人阅读

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  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 }

 


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