UVa 108-maximum Sum (maximum consecutive subsequence)

Title 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's simple-to-solve in one dimension was often much more difficult-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 the "the elements in" that rectangle. The problem the sub-rectangle with the largest sum are 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 an integer N on a line by itself indicating the size of the square and the Dimen Sional Array. This was followed by integers separated by white-space (newlines and spaces). These integers make to the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers o n the second row, left-to-right, etc.). N may be as large as 100. The numbers in the array would 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

Problem Solving Ideas:

Test instructions: The n*n matrix is given, and the maximum value of the matrix is obtained.

The application of the maximal continuous subsequence sequence is one-dimensional, the matrix is two-dimensional, so we can convert the matrix to one-dimensional to calculate.

That is, the enumeration matrix of successive lines of the merge, so that the conversion to one-dimensional, and then the maximum sub-sequence of the algorithm to seek, update the maximum value can be.

Code:

1#include <bits/stdc++.h>2 3 using namespacestd;4 5 inttable[ -][ -];6 intsum[ -];7 intN;8 9 intmax_continuous_sum ()Ten { One     intmaxs=0, s=0; A      for(intI=0; i<n; i++) -     { -         if(s>=0) s+=Sum[i]; the         Elses=Sum[i]; -Maxs = Maxs>s?maxs:s; -     } -     returnMaxs; + } - intMain () + { ACIN >>N; at     intmaxsum=0; -     inttmp; -      for(intI=0; i<n; i++) -     { -          for(intj=0; j<n; J + +) -         { inCIN >>Table[i][j]; -sum[j]=Table[i][j]; to         } +TMP =max_continuous_sum (); -Maxsum = maxsum>tmp?maxsum:tmp; the          for(intj=i-1; j>=0; j--) *         { $              for(intk=0; k<n; k++)Panax Notoginsengsum[k]+=Table[j][k]; -TMP =max_continuous_sum (); theMaxsum = maxsum>tmp?maxsum:tmp; +         } A     } thecout << maxsum <<Endl; +     return 0; -}

UVa 108-maximum Sum (maximum consecutive subsequence)