POJ1050(dp)

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46788		Accepted: 24774

Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*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 * 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

 最大子段和的二维版本，把第i行到第j行合并成一行，做法就和一维的一样了，只要枚举i和j，找出最大值即为答案。

//2016.8.21
#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N = 105;
const int inf = 0x3f3f3f3f;
int a[N][N];

int main()
{
    int n, tmp;
    while(scanf("%d", &n)!=EOF)
    {
        for(int i = 0; i < n; i++)
              for(int j = 0; j < n; j++)
                  scanf("%d", &a[i][j]);
        
        int ans = -inf;
        for(int i = 0; i < n-1; i++)
        {
//把第j行合并到第i行，求出第i行到第j行的最大子段和**************
            for(int j = i; j < n; j++)
            {
                tmp = 0;
                for(int k = 0; k < n; k++)
                {
                    if(j > i) a[i][k]+=a[j][k];//把矩阵合并为一维的数组
                    if(tmp > 0) tmp += a[i][k];
                    else tmp = a[i][k];
                    ans = max(ans, tmp);
                }
            }
//**************************************************************            
        }

        cout<<ans<<endl;
    }

    return 0;
}