[动态规划]UVA108 - Maximum Sum

 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

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