To The Max（最大子矩阵问题）

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9268 Accepted Submission(s): 4495

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

这个是动态规划的最大子矩阵，跟最大子段的区别就是它是二维

的！所以要把它压缩成一维！这里的意思就是先算出每一列，再

算出每一行

这一题就是将一维的最大字段和扩展到二维，在一维的求最大字段和的过程中是这样操作的：

int max_sum(int n)
{
    int i, j, sum = 0, max = -10000;
    for(i = 1; i <= n; i++)
    {
        if(sum < 0)
            sum = 0;
        sum += a[i];
        if(sum > max)
            max = sum;
    }
    return sum;
}

扩展到二维的时候也是同样的方法，不过需要将二维压缩成一维，所以我们要将数据做一下处理，使得map[i][j]从表示第i行第j个元素变成表示第i行前j个元素和，这样map[k][j]-map[k][i]就可以表示第k行从i->j列的元素和。只要比一维多两层循环枚举i和j就行了。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int map[105][105];
int maxs;
int main()
{
    int n,i,j,v,k,sum;
    while(cin>>n)
    {
        memset(map,0,sizeof(map));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                cin>>v;
                map[i][j]=map[i][j-1]+v;
            }
        }//这个的每一列的每一个数字是前面的数字之和！这样就确保了每一列是线性的！
        maxs=-100000;
        for(i=1;i<=n;i++)
        {
            for(j=i;j<=n;j++)
            {
                sum=0;
                for(k=1;k<=n;k++)
                {
                    sum+=map[k][j]-map[k][i-1];//每一行再进行线性就可以实现二维数组的动态规划！
                    if(sum<0)
                    {
                        sum=0;
                    }
                    if(sum>maxs)
                    {
                        maxs=sum;
                    }
                }
            }
        }
        cout<<maxs<<endl;
    }
}