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.
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
