Hdu 1081 长方形列举To The Max

Hdu 1081 长方形列举To The Max



Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7349 Accepted Submission(s): 3554





Problem Description

Given a two-dimensional array of positiveand negative integers, a sub-rectangle is any contiguous sub-array of size 1 x1 or greater located within the whole array. The sum of a rectangle is the sumof all the elements in that rectangle. In this problem the
sub-rectangle withthe largest sum is referred to as the maximal sub-rectangle.



As an example, the maximal sub-rectangle ofthe 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 ofintegers. The input begins with a single positive integer N on a line byitself, indicating the size of the square two-dimensional array. This isfollowed by N 2 integers separated by whitespace (spaces and newlines).
Theseare the N 2 integers of the array, presented in row-major order. That is, allnumbers 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 bein the range [-127,127].





Output

Output the sum of the maximalsub-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



题解：

长方形列举，sum[i][j]记录从a[i][1]到a[i][j]的和，所以在下面列举长方形时，一开始是从sum[1][1]列举起，宽度一定，先是长度的增加，如果面积小于0，则没有必要加到下面的面积了，重新计算宽度为一的长方形的面积。面积大于0，则往下加。然后增加宽度，先计算宽为1到i的，然后计算2-i的。而内层循环则计算宽度一定而长度变化的长方形的面积。面积大于0才加入下面的面积计算。在算的过程中求出最大的面积。三层循环。

源代码:

#include
<iostream>
#include
<stdio.h>
#include
<string.h>
using namespace std;

int main()
{
intsum[105][105];
intn,a;
while(cin>>n)
{
memset(sum,0,sizeof(sum));
for(int i = 1;i <= n;i++)
for(int j = 1;j <= n;j++)
{
scanf("%d",&a);
sum[i][j] = sum[i][j-1]+a;
}

intans = -1;
intmx = -0x3f3f3f3f;
for(int i = 1;i <= n;i++)
for(int j = 1;j <= i;j++)
{
ans = -1;
for(int k = 1;k <= n;k++)
{
if(ans< 0)
ans = sum[k][i] -sum[k][j-1];
else
ans += sum[k][i] -sum[k][j-1];

if(ans> mx)
mx = ans;
}
}
printf("%d\n",mx);
}
return0;
}