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To The Max (动态规划//最大连续子序列和)

2018-06-28 14:08:54         来源:duanghaha的博客  
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To The Max (动态规划//最大连续子序列和) 。

Problem DescriptionGiven a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. 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.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.
?
InputThe input consists of an N x N 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 N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in 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].
?
OutputOutput 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

降维运算,我们可以先把从i到j行的每一列加起来求最大连续子序列和,然后求出最大值即可(注意多组输入);

#include
#include
#include
using namespace std;
typedef long long ll;
#define MAXN 1000
const int INF=1<<29;
int a[MAXN][MAXN];
int b[1000050];
int main()
{
	int n;
	while(cin>>n)
	{
		for(int i=0;i>a[i][j];
			}
		}
		int ans=-INF;
		for(int i=0;i0) sum+=b[k];
					else sum=b[k];
					if(sum>ans) ans=sum;
				}
			}
		}
		cout<
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