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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15 题目的大意就是说在给出的n*n个数组成的矩形中饭找出一个小矩形，使得这些数的和最大。 这道题可以借鉴求最大和的连续子序列的方法来做； 在i，j行之间求出每一列这之间的和，那么这个和的最大连续子序列和就是所求 如样例：  0 -2 -7  0  9  2 -6  2 -4  1 -4  1 -1  8  0 -2 假如在第一行到第三行之间求每一列的和那就是 -13 1 -17 3 这个序列的最大连续区间的和就是举行的高为1-3行之间的最大矩形里的和 那么我们就只需要枚举所有行的区间就行了，连续区间最大和的球阀复杂度是O(n)，枚举所有行是O(n^2)，所以总的复杂度就是O(n^3)。n<=100所以不会超时

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #define MAX(a,b) (a) > (b)? (a):(b) 6 #define MIN(a,b) (a) < (b)? (a):(b) 7 #define mem(a) memset(a,0,sizeof(a)) 8 #define INF 1000000007 9 #define MAXN 10510 #define MAXC 100511 using namespace std;12 13 int N,ma[105][105];14 int sumU[105][105],sum[105];15 int main()16 {17     while(~scanf("%d",&N))18     {19         mem(sumU);20         int i,j,ans = -INF;21         for(i=1;i<=N;i++)for(j=1;j<=N;j++)22         {23             scanf("%d",&ma[i][j]);24             sumU[i][j]=sumU[i-1][j]+ma[i][j];25         }26         int Top,Down;27         for(Top=0;Top
         
          0?0:(sumU[Down][1]-sumU[Top][1]);34                 for(i=2;i<=N;i++)35                 {36                     if(min<0)min=(sumU[Down][i]-sumU[Top][i]);37                     else min+=(sumU[Down][i]-sumU[Top][i]);38                     if(max