To The Max（最大子矩阵问题）

深藏阁楼爱情的钟 2022-08-05 05:05 178阅读 0赞

# 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;
        }
    }