最大子矩阵和（To the max） 杭电1081

To the Max

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.

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

题意：

给 N*N 的矩阵，找出某一个子矩阵，将其所有元素相加求和，求最大的和

即找一个最大的子矩阵和

例如：

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.

思路：

二维的直接求解太麻烦，我们转化成一维来求解

子矩阵也可以看成是原矩阵的某几行和某几列的和

上述例子可以看成是 第2行到第四行每一列相加求和，然后转化成一维的求一个最大字段和

比如：将第2行到第4行每一列相加求和得到： 4 11 -10 1

对这个一维数组求一个最大字段和结果是 15

因此：

我们可以遍历行与行的组合，对每一列求和，然后求最大字段和

行与行的组合关系有：1  1-2  1-3  1-4

                                     2  2-3  2-4  

                                     3  3-4

                                     4

因此遍历这几种关系队列求和即可

解决方法：

① 用三维数组sum[i][j][k]  分别代表 从第 i 行到第 j 行的第 k 列的和

② 对①中求得的结果进行最大字段和，我用的是动归

CODE： 

#include 
#include 
#include 
#include 
#include 
#define memset(a,n) memset(a,n,sizeof(a))
const int MAXX = 110;
using namespace std;
int sum[MAXX][MAXX][MAXX];
int dp[MAXX];
int n;
int maxSum(int r,int c)
{
    int maxx = 0;
    for(int i=0; i maxx)
            maxx = dp[i];
    }
    return maxx;
}
int main()
{
    int maxx,maxx1,maxx2;
    int a[MAXX][MAXX];
    memset(a,0);
    memset(sum,0);
    cin >> n;
    for(int i=0; i> a[i][j];
    maxx = -1;
    for(int i=0; i