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
解题思路:
题意:给出n*n的矩阵,求出里面子矩阵的和的最大值。
最大连续子序列的应用,序列是一维的,矩阵是二维的,所以我们可以把矩阵转换为一维的来算。
也就是枚举矩阵的连续几行的合并,这样就转换为一维的了,再用最大子序列的算法去求,更新最大值就可以了。
代码:
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 }