To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7484    Accepted Submission(s): 3625

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
 
 
/*首先 , 读数据的时候 a[i][j] 为前 i 行数据的第 j 个数的和 , 当求第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵的时候 , 就以 p 行和 q 行之间的同一列的数字的和作为一个数列的元素 , 然后 DP 求这个数列的最大和连续子序列 , 就是第 p 行到第 q 行之间宽为 p-q+1 的最大和矩阵 . */
代码如下：
#include <iostream>
#include<string.h>
using namespace std;
int f[101][101];
int main()
{
    int n,i,j,sum,x,y,max;
    while(cin>>n)
    {
    memset(f,0,sizeof(f));
    for(j=1;j<=n;j++)
       for(i=1;i<=n;i++)
        {
            cin>>f[j][i];
            f[j][i]+=f[j-1][i];
           // if(f[j][i]>sum) sum=f[j][i];
        }
    max=-127;
        for(i=1;i<=n;i++)      
         for(j=i;j<=n;j++)
          {
              sum=0;
              for(y=1;y<=n;y++)
              {
                  sum+=f[j][y]-f[i-1][y];
                  if(sum>max) max=sum;
                  if(sum<0) sum=0;
              }
          }
        cout<<max<<endl;
    }
    return 0;
}