UV 108, uva108

UV 108, uva108

Source: https://uva.onlinejudge.org/index.php? Option = com_onlinejudge & Itemid = 8 & category = 3 & page = show_problem & problem = 44

 

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-dimenstmarray 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 asMaximal 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 integerNOn a line by itself indicating the size of the square two dimen1_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 .).NMay 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

Solution:

Question: Give a matrix of n * n and find the maximum value of the sum of the Child matrices in it.

The maximum continuous subsequence is applied. The sequence is one-dimensional and the matrix is two-dimensional. Therefore, we can convert the matrix into one-dimensional computation.

That is, the merge of several consecutive rows of the enumeration matrix, which is converted to one-dimensional. Then, we can use the maximum subsequence algorithm to update the maximum value.

Code:

 1 #include <bits/stdc++.h> 2  3 using namespace std; 4  5 int table[100][100]; 6 int sum[100]; 7 int N; 8  9 int max_continuous_sum()10 {11     int maxs=0,s=0;12     for(int i=0; i<N; i++)13     {14         if(s>=0) s+=sum[i];15         else s=sum[i];16         maxs = maxs>s ? maxs : s;17     }18     return maxs;19 }20 int main()21 {22     cin >> N;23     int maxsum=0;24     int tmp;25     for(int i=0; i<N; i++)26     {27         for(int j=0; j<N; j++)28         {29             cin >> table[i][j];30             sum[j]=table[i][j];31         }32         tmp = max_continuous_sum();33         maxsum = maxsum>tmp ? maxsum : tmp;34         for(int j=i-1; j>=0; j--)35         {36             for(int k=0; k<N; k++)37                 sum[k]+=table[j][k];38             tmp = max_continuous_sum();39             maxsum = maxsum>tmp ? maxsum : tmp;40         }41     }42     cout << maxsum << endl;43     return 0;44 }