uva 10827 - Maximum sum on a torus

Problem H Maximum sum on a torus Input: Standard Input
Output: Standard Output
A grid that wraps both horizontally and vertically is called a torus. Given a torus where each cell contains an integer, determine the sub-rectangle with the largest sum. The sum of a sub-rectangle is the sum of all the elements in that rectangle. The grid below shows a torus where the maximum sub-rectangle has been shaded.

-1
0
0
-4


-2
-3



-1

0

-2

-3

-3



-4
Input
The first line in the input contains the number of test cases (at most 18). Each case starts with an integer N (1≤N≤75) specifying the size of the torus (always square). Then follows N lines describing the torus, each line containing N integers between -100 and 100, inclusive.

Output


For each test case, output a line containing a single integer: the maximum sum of a sub-rectangle within the torus.

Sample Input Output for Sample Input

2
5
1 -1 0 0 -4
2 3 -2 -3 2
4 1 -1 5 0
3 -2 1 -3 2
-3 2 4 1 -4
3
1 2 3
4 5 6
7 8 9
15
45
Problem setter: Jimmy Mårdell
Special Thanks: Derek Kisman, Md. Kamruzzaman uva108 강화판, 직사각형은 순환적이다. 세 개를 추가한다. 직감은 우리가 먼저 나타날 수 있는 모든 직사각형을 하나하나 열거한 다음에 108의 코드를 씌우고 과감한 TLD를 써서 모든 줄의 조합을 약간 최적화시키거나 가장 큰 서열의 시작점을 하나하나 계산하는 것을 생각하게 한다. 시간이 지나갔다.
#include<stdio.h>
#include<string.h>
#define inf -9999999
int b[200],n;
int count()
{
 int i,j,f=1,max=inf,sum;
 for (i=1;i<=n;i++)
 {
  if  (b[i]>max) max=b[i];
  if (b[i]>=0)  f=0;
 }
 if (f) return max;
 for (j=1;j<=n;j++)
 {
  sum=0;
  for (i=0;i<n;i++)
  {
   sum=sum+b[i+j];
   if (sum>max) max=sum;
   if (sum<0) sum=0;
  }
 }
 return max;
}
int main()
{
 int T,i,j,p,q,k,t,a[200][200],max,ans,m;
 scanf("%d
",&T); while (T--) { scanf("%d",&n); for (i=1;i<=n;i++) for (j=1;j<=n;j++) { scanf("%d",&a[i][j]); a[i+n][j]=a[i][j]; a[i][j+n]=a[i][j]; a[i+n][j+n]=a[i][j]; } max=inf; for (k=0;k<n;k++) { for (i=1;i<=2*n;i++) { if (i+k<=2*n) { memset(b,0,sizeof(b)); for (j=i;j<=i+k;j++) for (t=1;t<=2*n;t++) b[t]+=a[j][t]; ans=count(); if (ans>max) max=ans; } } } printf("%d
",max); } return 0; }

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