POJ 1050 To the Max

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36627		Accepted: 19262

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
题目大意：最大子矩阵和问题。
解题方法：动态规划。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

int maze[105][105];
int Subsum[105];

int GetSubsum(int n)
{
    int sum = 0;
    int Maxsum = -10000000;
    for (int i = 0; i < n; i++)
    {
        if (sum >= 0)
        {
            sum += Subsum[i];
        }
        else
        {
            sum = Subsum[i];
        }
        Maxsum = max(Maxsum, sum);
    }
    return Maxsum;
}

int main()
{
    int n;
    scanf("%d", &n);
    int Maxsum = -10000000;
    for (int i = 0; i< n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            scanf("%d", &maze[i][j]);
        }
    }
    for (int i = 0; i < n; i++)
    {
        memset(Subsum, 0, sizeof(Subsum));
        for (int j = i; j < n; j++)
        {
            for (int k = 0; k < n; k++)
            {
                Subsum[k] += maze[k][j];
            }
            Maxsum = max(Maxsum, GetSubsum(n));
        }
    }
    printf("%d\n", Maxsum);
    return 0;
}