To The Max-蒲公英云

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9296 Accepted Submission(s): 4507

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

Source

Greater New York 2001

这一题原来是先把每一行的列算成现线性的，到最后再进行行的

动态规划！现在不一样了，我用矩形思想再写一次发现AC了！

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int a[105][105];
int main()
{
    int n,i,j,k,l,maxs,ans,f1,f2;
    while(cin>>n)
    {
        maxs=-10000;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                cin>>a[i][j];
                a[i][j]=a[i][j]+a[i-1][j]+a[i][j-1]-a[i-1][j-1];
            }
        }
      /*  for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                cout<<a[i][j]<<" ";
            }
            cout<<endl;
        }*/
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                for(k=1;k<=i;k++)
                {
                    for(l=1;l<=j;l++)
                    {
                        f1=i-k+1;f2=j-l+1;//这个是为了求出矩形的长和宽1
                        ans=a[i][j]-a[i][j-f2]-a[i-f1][j]+a[i-f1][j-f2];
                        maxs=max(ans,maxs);
                    }
                }
            }
        }
        cout<<maxs<<endl;
    }
}