POJ-1050 To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 56579		Accepted: 29921

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

Source

Greater New York 2001

ID-OJ:
POJ-1050

author:
Caution_X

DATE of submission:
20,191,117

Tags:
two-dimensional array prefix and

description modelling:
calculating a two-dimensional array of weights and the sub-matrix largest

major steps to solve it:
enumeration i (i∈ [1 , n-]) to j (j∈ [i + 1, n]) for each one-dimensional array of columns and rows of the elements, and the maximum value is calculated and updated answer to the one-dimensional array of continuous sections right

warnings:
a two-dimensional array into a one-dimensional array to handle, reduce the time overhead

AC code:

#include <stdio.h>
 #define the MAXSIZE 101 // determined row and the largest sub-segment int MaxArray ( int n-, int arr_ []) 
{ int I, sum_ = 0 , max_ = 0 ;
     for (I = . 1 ; I <= n-; I ++ ) 
    { IF (sum_> 0 ) 
        { 
            sum_ + = arr_ [I]; 
        } the else 
        { 
            sum_ = arr_ [I]; 
        } IF (sum_> max_) 
        {



    
        
        
        
            max_ =         {sum_; 
        } 
    } 
    return max_; 
} 
// find the maximum and the sub-matrix. 
int MaxMatrix ( int n-, int ARR _ [] [the MAXSIZE]) 
{ 
    int max_ = arr_ [ . 1 ] [ . 1 ];
     int sum_;
     int I, J, K;
     int temp_arr [the MAXSIZE]; 
     for (I = . 1 ; I < n =; I ++) // starting from the first row, until the n-th row 
    {
         for (J = . 1 ; J <= n; J ++) // only changes the start line, temp_arr initializing the array before 

            temp_arr [J] = 0 ; 
        }         
        for (J = i; J <= n; J ++) // from the i-th row to the n-th row 
        {
             for (K = . 1 ; K <= n; K ++ )   
            { 
                temp_arr [K] + arr_ = [J] [k]; // temp_arr [k] represents the sum of the k-th column of the i-th row to the n-th row. 
            }          
            Sum_ = MaxArray (n-, temp_arr); // determine the largest sub-segment row and        
            IF (sum_> max_) 
            { 
                max_ = sum_; 
            } 
        } 
    } 
    return max_; 
} 
int main()
{
    int n;
    int i, j;
    int arr_[MAXSIZE][MAXSIZE];
    int max_;
    while (~scanf("%d", &n)) //多组测试。 相当于 scanf("%d", &n) != EOF
    {
        for (i=1; i<=n; i++)
        {
            for (j=1; j<=n; j++)
            {
                scanf("%d", &arr_[i][j]);
            }
        }
        max_ = MaxMatrix(n, arr_);
        printf("%d\n", max_);
    }
    return 0;
}