To the Max
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:
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
Source
思路：经典dp（yan 教的）
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+,M=1e6+,inf=1e9+;
int a[N][N];
int sum[N][N];
int num[N];
int main()
{
int x,y,z,i,t;
while(~scanf("%d",&x))
{
memset(sum,,sizeof(sum));
for(i=;i<=x;i++)
for(t=;t<=x;t++)
scanf("%d",&a[i][t]);
for(i=;i<=x;i++)
for(t=;t<=x;t++)
sum[t][i]=sum[t-][i]+a[t][i];
int maxx=-inf;
for(i=;i<=x;i++)
{
for(t=i;t<=x;t++)
{
for(int j=;j<=x;j++)
num[j]=sum[t][j]-sum[i-][j];
int sum=;
for(int j=;j<=x;j++)
{
sum+=num[j];
maxx=max(maxx,sum);
if(sum<)
sum=;
}
}
}
printf("%d\n",maxx);
}
return ;
}