杭电1081 最大子矩阵和

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11573    Accepted Submission(s): 5578

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

  
  

   
   15
  
  

最大子矩阵和模板题。

可以转化成最大子序列和来做，用i,k表示行数，将第i行到第k行每一列数依次相加，存储在数组sum[]中，求出sum[]的最大子序列和。

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,a[105][105],dp[105],sum[105],max1;
int main()
{
    //freopen("D://in.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        max1=-0xFFFFF;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&a[i][j]);
        for(int i=0;i<n;i++)
        {
            memset(sum,0,sizeof(sum));
            for(int k=i;k<n;k++)
            {
                for(int j=0;j<n;j++)
                {
                    sum[j]+=a[k][j];
                    dp[j]=-0xFFFFF;
                }
                dp[0]=sum[0];
                max1=max(max1,dp[0]);
                for(int j=1;j<n;j++)
                {
                    dp[j]=max(sum[j],dp[j-1]+sum[j]);
                    max1=max(max1,dp[j]);
                }
            }
        }
        printf("%d\n",max1);
    }
    return 0;
}