题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1081

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8839    Accepted Submission(s):
4281

Problem Description

Given a two-dimensional array of positive and negative
integers, a sub-rectangle is any contiguous sub-array of size 1 x 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 x 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

题目大意：在一个数的矩阵里面找一个和最大矩阵。

题目思路:化二维为一维，转化成1003即可。先求一列一列的和，这样就转化成了一维了。举个例子，我先计算第一行得到dp[0]=0dp[1]=-2,dp[2]=-7,dp[3]=0;这些就是所谓的和，然后在用1003的方法做。那么第二行，就会更新，dp[0]=9,dp[1]=0,dp[2]=-13,dp[3]=2;这些依旧是和，这也就是转换成了一维数组，用1003的方法解决，以此类推~~~

详见代码。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4
 5 using namespace std;
 6
 7 int dp[110];
 8 int num[110][110];
 9
10 int main ()
11 {
12     int n;
13     while (~scanf("%d",&n))
14     {
15         for (int i=0; i<n; i++)
16             for (int j=0; j<n; j++)
17                 scanf("%d",&num[i][j]);
18         int ans=-99999;
19         for (int i=0; i<n; i++)
20         {
21             memset(dp,0,sizeof(dp));
22             for (int j=i; j<n; j++)
23             {
24                 int Max=-1;
25                 for (int k=0; k<n; k++)
26                 {
27                     dp[k]+=num[j][k];       //先计算出所有的dp和
28                 }
29                 for (int k=0; k<n; k++)     //1003的做法，代码类似
30                 {
31                     if (Max+dp[k]<dp[k])
32                         Max=dp[k];
33                     else
34                         Max=Max+dp[k];
35                     if (ans<Max)              //不断更新最大值
36                     {
37                         ans=Max;
38                         //cout<<j<<" "<<k<<endl;
39                     }
40                 }
41             }
42         }
43         printf ("%d\n",ans);
44     }
45     return 0;
46 }