集训第五周动态规划 F题 最大子矩阵和

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

这道题其实就是求最大子段和，需要把原题数据变化一下，例如0 -2 -7 0   9 2 -6 2 -4 1 -4 1 -1 8 0 -2 这个矩阵我选择的是9 2-4 1-1 8那么我还可以把这个选择过程看待为求数组 4 11 -10 1 的最大子段和，很显然是选择4 11，答案为15那么4 11 -10 1是怎么来的呢，是我把2 3 4行数组组合成一个数组得来的那么这道题的解法就出来了，不断枚举行区间，得到一个新数组，然后求最大子段和

#include"iostream"
#include"cstring"
using namespace std;
const int maxn=110;
int b[maxn],a[maxn][maxn];
int n;

void Init()
{
    int t;
    memset(a,0,sizeof(a));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            cin>>t;
            a[i][j]=a[i-1][j]+t;
        }
    }
}

int main()
{
    while(cin>>n)
    {
     Init();
     int sum,ans,temp;
     sum=0;
     ans=-1000000000;
     int c=1;
     for(int i=1;i<=n;i++)
     for(int j=i;j<=n;j++)
     {
     sum=0;
     for(int k=1;k<=n;k++)
     {
         temp=a[j][k]-a[i-1][k];
         sum+=temp;
         if(sum>ans)
         {
             ans=sum;
         }
         if(sum<0)
         {
             sum=0;
         }
     }
     }
      cout<<ans<<endl;
    }
    return 0;
}