poj 1050 TO THE MAX

最新推荐文章于 2021-10-04 16:49:11 发布

原创最新推荐文章于 2021-10-04 16:49:11 发布 · 154 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#DP

本文介绍了一种求解二维数组中最大子矩阵和的算法，通过预处理得到每行的前缀和，再利用动态规划的方法枚举不同长度的子矩阵，最终找到具有最大和的子矩阵。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

Sample Output

思路：

这题就是求最大子矩阵，可以先把每行的前缀和求出来，然后枚举区间长度，然后列向进行最大区间和dp，总共需要运算 ∑n*(n-i)(1<=i<=n)次，复杂度o(n^3)。下面是代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set> 
using namespace std;
typedef long long ll;
typedef long double ld;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const int MAXN=101;
int sum[MAXN][MAXN];
int DP(int l,int n)//区间长度为l的dp
{ int maxx=-INF,dp=0;
  for(int i=0;i<=n-l;i++)
  { dp=0;
    for(int j=1;j<=n;j++)
    {if(dp>=0) dp=dp+sum[j][i+l]-sum[j][i];
     else dp=sum[j][i+l]-sum[j][i];
     if(dp>maxx) maxx=dp;
	}
  }
  return maxx;
}
int main()
{ int ans;
  int n,x;
  while(~scanf("%d",&n))
  { ans=-INF;
   memset(sum,0,sizeof(sum));
   for(int i=1;i<=n;i++)
   for(int j=1;j<=n;j++)
   {scanf("%d",&x);
    sum[i][j]=sum[i][j-1]+x;
   }
   for(int i=1;i<=n;i++) ans=max(ans,DP(i,n));
   printf("%d\n",ans);
  }
}