To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11009    Accepted Submission(s): 5278

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

40 -2 -7 0 9 2 -6 2-4 1 -4 1 -18 0 -2

 

Sample Output

15

 

#include <iostream>#include <cstring>using namespace std;int buf[105],n;int getmax(){    int Temp[105],max=127*(-127);    memset(Temp,0,sizeof(Temp));    for(int i=1; i<=n; i++)    {        Temp[i]=(Temp[i-1]>0 ? Temp[i-1] : 0 )+buf[i];        if(max<Temp[i])            max=Temp[i];    }    return max;}//求最大连续和int main(){    while(cin>>n)    {        int maxsum=-127*n;        int a[n][n];        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                cin>>a[i][j];            }        }        for(int i=0; i<n; i++)        {            for(int j=i; j<n; j++)            {                memset(buf,0,sizeof(buf));                for(int k=0; k<n; k++)                {                    for(int l=i; l<=j; l++)                    {                        buf[k]+=a[k][l];                    }                }                int d=getmax();                if(d>maxsum)                    maxsum=d;            }        }        cout<<maxsum<<endl;    }    return 0;}