HDU1081 To The Max (DP)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11302    Accepted Submission(s): 5447

[align=left]Problem Description[/align]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.
 [align=left]Input[/align]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].
 [align=left]Output[/align]Output the sum of the maximal sub-rectangle.
 [align=left]Sample Input[/align]

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2 [align=left]Sample Output[/align]

15 
[align=left]Source[/align]Greater New York 2001
int MaxSubArray(int *a,int n)
{
int Max=-65535;
int i,tmp=0;
for(i=0;i<n;i++)
{
if(tmp>0)tmp+=a[i];
else tmp=a[i];
if(tmp>Max) Max=tmp;
}
return Max;
}
/*求 n 行，m 列的矩阵的最大子矩阵和 */
int MaxSubMatrix(int n,int m)
{
int Max=-65535;
int i,j,k;
int sum;
int b[MAXN];
for(i=0;i<n;i++)
{
memset(b,0,sizeof(b));
for(j=i;j<n;j++)
{
for(k=0;k<m;k++) b[k]+=map[j][k];
sum=MaxSubArray(b,m);
if(sum>Max)Max=sum;
}
}
return Max;
}


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