Maximum Sum [UVA - 108 ]

最新推荐文章于 2021-07-29 15:39:30 发布

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本文介绍了一种寻找二维数组中最大子矩阵之和的有效算法。该算法通过预处理和动态规划的方法，在N×N的矩阵中找到具有最大和的子矩阵，并提供了完整的实现代码。

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the subrectangle with the largest sum is referred to as the maximal sub-rectangle.

A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array. 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-hand corner:

9 2

−4 1

−1 8

and has the 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 N2 integers separated by white-space (newlines and spaces). These N2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is the sum of the maximal sub-rectangle.

Sample Input

0 -2 -7 0

9 2 -6 2

-4 1 -4 1

-1 8 0 -2

Sample Output

题意：

求n*n的矩阵的最大子矩阵之和，用s数组存和。

代码：

#include<stdio.h>
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int a[105][105]= {0},s[105][105]= {0};
        int maxx=0,sum;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            {
                scanf("%d",&a[i][j]);
                s[i][j+1]=s[i][j]+a[i][j];
            }
        for(int i=0; i<n; i++)
            for(int j=i; j<=n; j++)
            {
                sum=0;
                for(int k=0; k<n; k++)
                {
                    sum=sum+s[k][j]-s[k][i];
                    if(sum<0)
                        sum=0;
                    else if(sum>maxx)     //更新
                        maxx=sum;
                }
            }
        printf("%d\n",maxx);
    }
    return 0;
}