hdu 1081 To The Max(暴力枚举+最大连续子数组和)

最新推荐文章于 2022-01-24 10:00:00 发布

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最新推荐文章于 2022-01-24 10:00:00 发布

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本文介绍了一种解决二维数组中寻找具有最大和的子矩阵的方法。通过枚举子矩阵的边界，并利用一维最大子数组和算法，可以高效地找到该子矩阵及其和。文章还提供了完整的代码实现。

题目链接

To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11502 Accepted Submission(s): 5540

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

Source

Greater New York 2001

注意这一题是多组数据，且n为0时结束输入！

此题的做法就是枚举子矩阵的上下行编号i，j，求出i行至j行每一列的和，然后对于这个数组求最大连续和，有点像扫描线。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=100+10;
const int inf=0x3f3f3f3f;
int a[maxn][maxn],sum[maxn],d[maxn];
int max_cnt(int* d,int n)             //求数组d最大连续和 
{
	int ans=-inf;
	int sum=0;
	for(int i=1;i<=n;i++)
	{
		sum+=d[i];
		ans=max(ans,sum);
		if(sum<0) sum=0;
	}
	return ans;
}

int main()
{
	int n;
	while(~scanf("%d",&n)&&n)
	{
		for(int i=1;i<=n;i++)
	     for(int j=1;j<=n;j++)
	     scanf("%d",&a[i][j]);
	    int ans=-inf;
	    for(int i=1;i<=n;i++)
	    {
		   memset(d,0,sizeof(d));
		   for(int j=i;j<=n;j++)
		   {
			 for(int k=1;k<=n;k++)
				d[k]+=a[j][k];
			ans=max(ans,max_cnt(d,n));
		   }
	    }
	    printf("%d\n",ans);
	}
}