POJ1050

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32838		Accepted: 17175

Description

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

Source

Greater New York 2001

 

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 

 5 using namespace std;

 6 

 7 int main()

 8 {

 9     int n;

10     int array[110][110];

11     int dp[110][110];

12     int sum, term, maxSum;

13 

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

15     {

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

17         {

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

19             {

20                 scanf("%d",&array[i][j]);

21             }

22         }

23         //dp

24         memset(dp,0,sizeof(dp));

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

26         {

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

28             {

29                 dp[i][j]=dp[i][j-1]+array[i][j];

30             }

31         }

32         maxSum=-65535;

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

34         {

35             for(int j=i; j<=n;j++)

36             {

37                 sum=-1;

38                 for(int k=1;k<=n;k++)

39                 {

40                     term=dp[k][j]-dp[k][i-1];

41                     if(sum<0)  //sum[i]=max{sum[i-1]+a[i],a[i]}

42                     {

43                         sum=term;

44                     }

45                     else

46                     {

47                         sum+=term;

48                     }

49                     if(sum>maxSum)

50                     {

51                         maxSum=sum;

52                     }

53                 }

54             }

55         }

56         printf("%d\n",maxSum);

57     }

58 

59     return 0;

60 }