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 最大字串和的升级对于数列A[1->n] dp[i]=dp[i-1]+a[i] dp[i-1]>0 || dp[i]=a[i] dp[i-1]<=0;现在将矩阵建立与最大字串和相关联的模型。矩阵压缩。滚动数组。 用s[k]表示第K列，第I行到第J行的和，把和看成一个数列中的一个AI即可。就转化成了最大字串和的问题。
1 #include"iostream"
2 #include"cstdio"
3 #include"cstring"
4 using namespace std;
5 const int ms=110;
6 int n,ans;
7 int matrix[ms][ms];
8 int s[ms];
9 int main()
10 {
11 int i,j,k,t;
12 while(scanf("%d",&n)!=EOF)
13 {
14 for(int i=1;i<=n;i++)
15 for(j=1;j<=n;j++)
16 scanf("%d",&matrix[i][j]);
17 ans=-0x7fffffff;
18 for(i=1;i<=n;i++) //从第I行出发的子矩阵
19 {
20 memset(s,0,sizeof(s));
21 for(j=i;j<=n;j++) //到达第J行的子矩阵
22 {
23 t=0;
24 for(k=1;k<=n;k++)//包含第K列 最大矩阵
25 {
26 s[k]+=matrix[j][k];
27 if(t<=0)
28 t=s[k];
29 else
30 t+=s[k];
31 if(t>ans)
32 ans=t;
33
34 }
35 }
36 }
37 printf("%d\n",ans);
38 }
39 return 0;
40 }