HDU 1081

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5027 Accepted Submission(s): 2388

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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<stdio.h>

 2 #include<string.h>

 3 int a[101][101],b[101];

 4 int subsequencesum(int a[],int n)

 5 {

 6     int sum=0,maxsum=-0x7fffff,i;

 7     for(i=1;i<=n;i++)

 8         if(maxsum<a[i])

 9             maxsum=a[i];

10     if(maxsum<=0)

11         return maxsum;

12     for(i=0;i<n;i++)

13     {

14         sum+=a[i+1];

15         if(sum>maxsum)

16             maxsum=sum;   

17         else

18             if(sum<0)

19                 sum=0;

20     }

21     return maxsum;

22 }                     

23 int main()

24 {

25          int n,max,ans,temp;

26         int i,j,k,T,m;

27          while(~scanf("%d",&n))//说的是一组，实际却是多组 

28          {

29             temp=ans=max=-0x7fffff;

30             for(i=1;i<=n;i++)

31                 for(j=1;j<=n;j++)

32                     scanf("%d",&a[i][j]);

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

34             {                         

35                     memset(b,0,sizeof(b));

36                     for(j=i;j<=n;j++)

37                     {

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

39                             {

40                                 b[k]+=a[j][k];

41                             }

42                             ans=subsequencesum(b,n);//按行枚举 ,猜测按列枚举时间差不多， 

43                             if(temp<ans) 

44                                 temp=ans;

45                     }

46             }

47             printf("%d\n",temp);

48         }

49         //while(1);

50         return 0;

51 }            

52         

53