题目大意：给出 n*n 的矩阵，找每隔数字之和最大的子矩阵，输出最大和。 
解题思路：枚举矩阵左上和右下的坐标，分别合并子矩阵的每列，使得二维转化为一维，然后利用连续子序列最大和去做就行。

Time limit    3000 ms

题目描述：

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.

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 subrectangle with the largest sum is referred to as the maximal sub-rectangle.

A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array. 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-hand corner:

                                  9      2

                                −4      1

                                −1      8

and has the 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 N2 integers separated by white-space (newlines and spaces). These N2 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].

Output

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

题目理解：即在矩形中得到一个小矩形中总和最大数，即遍历长宽高

#include<iostream>
#include<string.h>
#define inf 0x3f3f3f3f
using namespace std;
int n,dp[1010][1010],ans;
int main()
{while(cin>>n){for(int i=0; i<n; i++)for(int j=0; j<n; j++)cin>>dp[i][j];ans=dp[0][0];for(int i=0; i<n; i++)for(int j=0; j<n; j++)if(j>=i){int sum=0;for(int k=0; k<n; k++){int step=0;for(int u=i; u<=j; u++){step+=dp[u][k];}sum+=step;if(sum>ans)ans=sum;if(sum<0)sum=0;}}cout<<ans<<endl;}return 0;
}