题目：
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
思路：
把二维转换为一维，变成求一个序列的最大子串问题；
如-1 -2；-3 -4先变成-1 -2；-4 -6.
把第一行减第零行就是求矩阵第一行的最大值，第二行减第零行
就是求整个矩阵的最大值；第二行减第一行就是求矩阵第二行的
最大值！
代码：
1 #include<stdio.h>
2 #include<string.h>
3 #define inf -0xfffffff
4 //#include<stdlib.h>
5 /*#include<iostream>
6 #include<algorithm>
7 using namespace std;*/
8
9
10 int main(){
11 int n, i, j, value[101][101], dp[101][101], max, zmax, k, sum;
12 while(scanf("%d", &n) != EOF){
13 memset(dp, 0, sizeof(dp));
14 for(i = 1; i <= n; i ++){
15 for(j = 1; j <= n; j ++){
16 scanf("%d", &value[i][j]);
17 dp[i][j] = dp[i - 1][j] + value[i][j];
18 }
19 }
20 zmax = inf;
21 for(i = 1; i <= n; i ++){
22 for(j = i; j <= n; j ++){
23 max = inf;
24 sum = 0;
25 for(k = 1; k <= n; k ++){
26 sum += dp[j][k] - dp[i - 1][k];
27 if(sum > max){
28 max = sum;
29 }
30 if(sum < 0){
31 sum = 0;
32 }
33 }
34 if(max > zmax){
35 zmax = max;
36 }
37 }
38 }
39 printf("%d\n", zmax);
40 }
41 return 0;
42 }
奋进