POJ 1050 To the Max (最大子矩阵和)

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该博客介绍了如何解决编程在线判题POJ 1050中的问题，即找到一个矩阵中的最大子矩阵和。内容包括问题描述、输入输出格式示例以及可能的解题思路。

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43888		Accepted: 23267

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


思路：都知道最大子序列和的求法，这道题就是对最大子序列有一个改变，定义一个sum数组记录前几行数组的和
例如：
1 2 3 -1
4 6 -2 5
那么他们的sum值为：5 8 1 4，然后就按照求最大子序列和的方法对sum数组进行遍历
然后依次往下推，在求最大子矩阵和的时候，还要考虑到减去上面几行的情况
详情见代码：



ac代码：

16ms
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAX(a,b) a>b?a:b
#define MAXN 1010
#define INF 0xfffffff
using namespace std;
int num[MAXN][MAXN];
int sum[MAXN][MAXN];
int main()
{
	int i,j,k,q;
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			scanf("%d",&num[i][j]);
		}
		memset(sum,0,sizeof(sum));//初始化sum数组
		int max=0;
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				sum[i][j]=i==0?num[i][j]:(sum[i-1][j]+num[i][j]);//第一行的sum值为其本身，当然这也可以在输入的同时更新
			}
			for(k=-1;k<i;k++)//考虑减去前几行
			{
				int ans=0;
				if(k>=0)
				{
					for(q=0;q<n;q++)
					{
						ans+=(sum[i][q]-sum[k][q]);
						if(ans<0)
						{
							ans=0;
						}
						else
						{
							max=MAX(max,ans);
						}
				    }
			    }
			    else//一行都不减那就是其本身
			    {
					for(q=0;q<n;q++)
					{
						ans+=sum[i][q];
						if(ans<0)
						{
							ans=0;
						}
						else
						{
							max=MAX(max,ans);
						}
				    }
			    }
			}
		}
		printf("%d\n",max);
	}
	return 0;
}

将sum的赋值加在输入中
0ms：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAX(a,b) a>b?a:b
#define MAXN 1010
#define INF 0xfffffff
using namespace std;
int num[MAXN][MAXN];
int sum[MAXN][MAXN];
int main()
{
	int i,j,k,q;
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		memset(sum,0,sizeof(sum));
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{ 
			     scanf("%d",&num[i][j]);
			     sum[i][j]=i==0?num[i][j]:(sum[i-1][j]+num[i][j]); 
			} 
		}
		int max=0;
		for(i=0;i<n;i++)
		{
            for(k=-1;k<i;k++) 
            {  
                int ans=0;  
                if(k>=0)  
                {  
                    for(q=0;q<n;q++)  
                    {  
                        ans+=(sum[i][q]-sum[k][q]);  
                        if(ans<0)  
                        {  
                            ans=0;  
                        }  
                        else  
                        {  
                            max=MAX(max,ans);  
                        }  
                    }  
                }  
                else
                {  
                    for(q=0;q<n;q++)  
                    {  
                        ans+=sum[i][q];  
                        if(ans<0)  
                        {  
                            ans=0;  
                        }  
                        else  
                        {  
                            max=MAX(max,ans);  
                        }  
                    }  
                }  
            }
		}
		printf("%d\n",max);
	}
	return 0;
}