http://poj.org/problem?id=1050
方法：二维转一维
1、使用积累数组cumarr[1...n][1...n]，使得求任意块儿子矩阵和的复杂度为O(1)，cumarr[i][j]为cumarr[1...i][1...j]子矩阵的和，
通过cumarr[i][j]=cumarr[i-1][j]+cumarr[i][j-1]-cumarr[i-1][j-1]计算积累数组，复杂度为O(N*N),
2、枚举+dp
对矩阵的行枚举，确定矩阵区域的上下界imin和imax，此时就可以进行类似一维数组最大和的方法，每个枚举的矩阵区域负责度O(N)。
总复杂度为O(N*N)+(N*N)*O(N)=O(N^3)
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
#include <iostream>
3:
namespace std ;
5:
//只用1...n，以便和积累数组对应
int a[101][101];
int cumarr[101][101];
9:
void run1050()
11: {
int n,val ;
int i,j ;
int imin,imax ;
int maxendinghere,maxsofar ;
int tmp ;
17:
, &n );
19:
//积累数组边界值
for( i=0 ; i<=n ; ++i )
22: {
23: cumarr[0][i] = 0 ;
24: cumarr[i][0] = 0 ;
25: }
26:
//输入矩阵并求得积累数组
for( i=1 ; i<=n ; ++i )
29: {
for( j=1 ; j<=n ; ++j )
31: {
, &(a[i][j]) ) ;
33: cumarr[i][j] = cumarr[i-1][j] + cumarr[i][j-1] - cumarr[i-1][j-1] + a[i][j] ;
34: }
35: }
36:
//对矩阵的行枚举，确定矩阵区域的上下界imin和imax，
38: maxsofar = a[1][1] ;
for( imin=1 ; imin<=n ; ++imin )
40: {
for( imax=imin ; imax<= n ; ++imax )
42: {
//进行类似一维数组最大和的方法
44: maxendinghere = cumarr[imax][1] - cumarr[imin-1][1] ;
for( j = 2 ; j<=n ; ++j )
46: {
47: tmp = cumarr[imax][j] - cumarr[imin-1][j] - cumarr[imax][j-1] + cumarr[imin-1][j-1] ;
48: maxendinghere = std::max( maxendinghere+tmp , tmp ) ;
49: maxsofar = std::max( maxsofar , maxendinghere ) ;
50: }
51: }
52: }
53:
, maxsofar ) ;
55: }