To the Max
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 44723 Accepted: 23679
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
这道题目是hdu1003 的升级版，HDU 1003，是一维数组最长子段和的问题，这个题目扩展到二维，思路就是把二维转换成一维，
先求第一行最大子段和，再求第一行跟第二行合起来的最大子段和，再求第一行到第三行合起来的最大值，实际上就是把二维数组转换成一维的了，
#include <iostream>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <stdlib.h>
using namespace std;
int a[105][105];
int n;
int dp[105];
int b[105];
int sum;
int main()
{
while(scanf("%d",&n)!=EOF)
{
sum=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&a[i][j]);
}
}
for(int i=1;i<=n;i++)
{
memset(b,0,sizeof(b));
memset(dp,0,sizeof(dp));
for(int k=i;k<=n;k++)
{
for(int j=1;j<=n;j++)
{
b[j]+=a[k][j];
if(dp[j-1]>=0)
dp[j]=dp[j-1]+b[j];
else
dp[j]=b[j];
if(sum<dp[j])
sum=dp[j];
}
}
}
printf("%d\n",sum);
}
return 0;
}