【HDU 1081】To The Max (动态规划//最大连续子序列和)

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14522    Accepted Submission(s): 6829

Problem Description

Given 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.

 

Input

The 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].

 

Output

Output 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<iostream>
#include<algorithm>
#include<string.h>
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<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				cin>>a[i][j];
			}
		}
		int ans=-INF;
		for(int i=0;i<n;i++)
		{
			memset(b,0,sizeof(b));
			for(int j=i;j<n;j++)
			{
				int sum=0;
				for(int k=0;k<n;k++)
				{
					b[k]+=a[j][k];//求每列的和 
					if(sum>0) sum+=b[k];
					else sum=b[k];
					if(sum>ans) ans=sum;
				}
			}
		}
		cout<<ans<<endl;	
	}
	return 0;
}