Maximum Sum

Background

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.

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 sub-rectangle with the largest sum is referred to as the maximal sub-rectangle. A sub-rectangle is any contiguous sub-array of size  or greater located within the whole array. As an example, the maximal sub-rectangle of the array:

is in the lower-left-hand corner:

and has the sum of 15.

Input and Output

The input consists of an  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  integers separated by white-space (newlines and spaces). These 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].

The output is 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

题意：给出一个方阵，这个这个方阵里面元素和最大的一个矩阵。

思路：最大连续子序列的升级版本，把原来的一维矩阵变成了二维的矩阵，需要转换思路，把二维矩阵看成一维数组，这个一维数组的元素是二维数组中同一列的元素相加。首先思路是枚举这个矩阵大小，再扫描。

#include<iostream>
#include<cstring>

using namespace std;

int matrix[110][100],sum[110][110];

int main()
    {
        int n;
        cin>>n;
        int i,j,k;
        memset(matrix,0,sizeof(matrix));
        for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                    {
                        cin>>matrix[i][j];
                        matrix[i][j]=matrix[i][j]+matrix[i-1][j];
                    }
            }
        int maxsum=matrix[1][1];
        for(i=0;i<=n;i++)
            {
                for(j=i;j<=n;j++)
                    {
                        int sum=0;
                        for(k=1;k<=n;k++)
                            {
                                if(sum<0) sum=matrix[j][k]-matrix[i][k];
                                else if(i!=j) sum=matrix[j][k]-matrix[i][k]+sum;
                                if(sum>maxsum&&sum!=0) maxsum=sum;
                            }
                    }
            }
        cout<<maxsum<<endl;
        return 0;
    }

[动态规划]UVA108 - Maximum Sum,布布扣,bubuko.com

[动态规划]UVA108 - Maximum Sum

原文：http://blog.csdn.net/zju_ziqin/article/details/23470397