1050pku To the Max 解题报告

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

解题思路：这个和一维数组截取一个字段得到的最值是一一样的，只是一维变成二维的，所以要对 数据进行处理，使得像一维数据那样处理，用一个三维的数组来保存一个列里，从i行到j行的和，然 后接着就是像一维数组一样从1~n列，只是这个不是一个数组而已，加了很多数组，这样就可以取他们 中最大的最大即可。（动态规划） 
代码如下： 

Code:

1. #include<iostream>
2. usingnamespacestd;
3. constintMax(101);
4. intsum[Max][Max][Max];//用来保存在一个列里，从i行到j行的和
5. intmain()
6. {
7. intn;
8. while(cin>>n)
9. {
10. intdata[Max][Max];
11. for(inti=1;i<=n;i++)
12. {
13. for(intj=1;j<=n;j++)
14. {
15. cin>>data[i][j];
16. sum[i][i][j]=data[i][j];//先初始化为自己那一格的值
17. }
18. }
19. for(inti=1;i<=n;i++)
20. {
21. for(intj=i+1;j<=n;j++)
22. {
23. for(intk=1;k<=n;k++)
24. {
25. sum[i][j][k]=sum[i][j-1][k]+data[j][k];//计算出在k列从i行到j行的和
26. }
27. }
28. }
29. intMaxSum=-9999;//先让初始化他们的最大值
30. for(inti=1;i<=n;i++)
31. {
32. for(intj=i;j<=n;j++)
33. {
34. inttemp=0;
35. for(intk=1;k<=n;k++)//和一维数组的得最大的段和是一样的操作
36. {
37. if(temp>=0)
38. temp+=sum[i][j][k];
39. else
40. temp=sum[i][j][k];
41. if(temp>MaxSum)
42. MaxSum=temp;
43. }
44. }
45. }
46. cout<<MaxSum<<endl;
47. }
48. return0;
49. }