poj1050 二维最大字段和

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

其实就是暴力一个最大字段和。。
本身最大字段和是求一行的
这个要求一个最大的矩形

那么你就枚举每一个可能出现的行…、
比如1到3，1到4全都叠加完了成为一个新的行
对新的行最大字段和就可以了..

之前我竟然往区间dp上想..
有点傻逼

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int tu[101][101];
int main()
{
    int n;
    cin >> n;
    int da = -100000000;
    for (int a = 1; a <= n; a++)
    {
        int tem = -10000000;
        for (int b = 1; b <= n; b++)
        {
            scanf("%d", &tu[a][b]);
            if (tem > 0)tem += tu[a][b];
            else tem = tu[a][b];
            da = max(da, tem);
        }
    }
    for (int a = 1; a <= n; a++)
    {
        for (int b = a + 1; b <= n; b++)
        {
            int tem = -1000000;
            for (int c = 1; c <= n; c++)
            {
                tu[a][c] += tu[b][c];
                if (tem > 0)tem += tu[a][c];
                else tem = tu[a][c];
                da = max(tem, da);
            }
        }
    }
    cout << da;
}