ACM学习-POJ-1050-To the Max

爱被打了一巴掌 2022-10-12 13:59 2阅读 0赞

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To the Max

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;37462</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;19724</td> 
  </tr> 
 </tbody> 
</table>

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

Source

[Greater New York 2001][]

**题目要求：求出最大子矩阵。**

**题目分析：**

**首先，如果遍历，找到所有的子矩阵，解出这道题是没问题的，只是在POJ上，必然会超时。**

**所以我考虑用动态规划来解决。即把这问题划分为更小的问题。**

**现在考虑把这个二维数组拆分，每行每行的先计算。**

**就把问题化解成求一维数组中子段和最大。 类似求最长字符串的问题。**

**然后累加比较总和。即可。**

**详细说明如下：**

1、首先考虑一维的最大子段和问题，给出一个序列a\[0\]，a\[1\]，a\[2\]...a\[n\]，求出连续的一段，使其总和最大。

a\[i\]表示第i个元素  
dp\[i\]表示以a\[i\]结尾的最大子段和

dp\[i\] = max\{a\[i\], dp\[i-1\] + a\[i\]\}

解释一下方程:

如果dp\[i-1\] > 0，则 dp\[i\] = dp\[i-1\] + a\[i\]  
如果dp\[i-1\] < 0，则 dp\[i\] = a\[i\]

因为不用记录位置信息，所以dp\[\]可以用一个变量dp代替:

如果dp > 0，则dp += a\[i\]  
如果dp < 0，则dp = a\[i\]

2、考虑二维的最大子矩阵问题

我们可以利用矩阵压缩把二维的问题转化为一维的最大子段和问题。因为是矩阵和，所以我们可以把这个矩形的高压缩成1，用加法就行了。

恩，其实这个需要自己画图理解，我的注释里写得很详细了，自己看吧。

枚举求的结果矩阵的行号范围，从(1,1), (1,2), ...到(1,n),然后再求(2,2),(2,3),...(2,n)的最大值，直至到(m,n)。

**下面给出AC代码。**

#include <stdio.h>
    #define MAXSIZE 101
    
    
    //求出一行中最大的子段和
    int MaxArray( int n, int arr_[])
    {
        int i, sum_ = 0, max_ = 0;
        for (i=1; i<=n; i++)
        {
            if (sum_>0)
            {
                sum_ += arr_[i];
            }
            else
            {
                sum_ = arr_[i];
            }
            if (sum_>max_)
            {
                max_ = sum_;
            }
        }
        return max_;
    }
    
    
    //求出最大子矩阵和。
    int MaxMatrix( int n, int arr_[][MAXSIZE])
    {
        int max_ = arr_[1][1];
        int sum_;
        int i, j, k;
        int temp_arr[MAXSIZE]; 
        for (i=1; i<=n; i++) //从第一行开始，直到第n行
        {
            for (j=1; j<=n; j++) //只有起始行改变，temp_arr数组才初始化
            {
                temp_arr[j] = 0;
            }
            
            for (j=i; j<=n; j++) //从i行到第n行
            {
                for (k=1; k<=n; k++)  
                {
                    temp_arr[k] += arr_[j][k]; //temp_arr[k] 表示从第i行到第n行中第k列的总和。
                }
                
                sum_ = MaxArray(n, temp_arr); //求出该行中最大的子段和
                
                if (sum_ > max_)
                {
                    max_ = sum_;
                }
            }
        }
        
        return max_;
    }
    
    int main()
    {
        int n;
        int i, j;
        int arr_[MAXSIZE][MAXSIZE];
        int max_;
        while (~scanf("%d", &n)) //多组测试。 相当于 scanf("%d", &n) != EOF
        {
            for (i=1; i<=n; i++)
            {
                for (j=1; j<=n; j++)
                {
                    scanf("%d", &arr_[i][j]);
                }
            }
            max_ = MaxMatrix(n, arr_);
            printf("%d\n", max_);
        }
        
        return 0;
    }

[Greater New York 2001]: http://poj.org/searchproblem?field=source&key=Greater+New+York+2001