Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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

   

這道題實際上是一維最大子序列和的問題在二維矩陣上的擴充套件。回憶一下，最大子序列和是求一維陣列中連續子序列最大和的問題，那麼怎麼向二維擴充套件呢？實際上，子矩陣在兩個方向上也是連續的。所以可以將這二維的問題轉化為一維的，就是可以將子矩陣每一列相加就會得到一個一維的陣列，然後可以用一維最大子序列和的方法來求解。剩下的就是要按行遍歷所有可能的子矩陣，比較找出其中的最大值。當然，思路是別人的，但是，程式碼是自己寫的。

#include <iostream>
#include<cstdlib>

using namespace std;
int maxsubarray(int *a,int n)
{
    int maxn=-1;
    int sum=0;
    for(int i=0;i<n;i++)
    {
        sum+=a[i];
        if(sum>maxn)
            maxn=sum;
        if(sum<0)
            sum=0;
    }
    return maxn;
}
int tothemax(int a[][101],int n)
{
    int t[101];
    int maxsum=0,maxn;
    for(int i=0;i<n;i++)
        for(int j=0;j<=i;j++)
        {
            for(int k=0;k<n;k++)
            {
                t[k]=0;
                for(int p=j;p<=i;p++)
                   t[k]+=a[p][k];
            }
            maxn=maxsubarray(t,n);
            if(maxsum<maxn) maxsum=maxn;
        }
    return maxsum;
}
int main()
{
    int i,j,maxsum;
    int N;
    cin>>N;
    int a[101][101];

    for(i=0;i<N;i++)
        for(j=0;j<N;j++)
           cin>>a[i][j];
    maxsum=tothemax(a,N);
    cout<<maxsum<<endl;
    return 0;
}