To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 28700 | Accepted: 14919 |
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:
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].
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
http://poj.org/problem?id=1050
很典型的动态规划的题目，这道题我是用了枚举+动态规划的方法完成的，先枚举目标矩阵的起始和结束行，再计算出每一列的数据的和，变化为求最大子段和，再利用动态规划求最大子段和，只需求出各种情况下的最大子段和即可。
#include <iostream>
using namespace std;
#define MAX 100
int main()
{
int n,i,j,k,sum=-12800;
int squma[MAX][MAX]={0};
int linesum[MAX][MAX]={0};
cin>>n;
for (i=0;i<n;++i)
for (j=0;j<n;++j)
cin>>squma[i][j];
for (i=0;i<n;++i)
for (j=i;j<n;++j)
{
int b=0;
for (k=0;k<n;++k)
{
linesum[i][k]+=squma[j][k];
if (b>0)
b+=linesum[i][k];
else
b=linesum[i][k];
if (b>sum)
sum=b;
}
}
cout<<sum;
return 0;
}