To The Max

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12471 Accepted Submission(s): 5985

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

#include <iostream>

#include <algorithm>

#include <vector>

#include<string.h>

using namespace std;

int a[][];

int main()

{

    int n,m;

    cin>>n>>m;

    for(int i=;i<n;i++)

    {

        for(int j=;j<m;j++)

        {

            cin>>a[i][j];

            a[i][j]+=a[i-][j];

        }

    }

    int dp[];

    memset(dp,,sizeof(dp));

    int ans=-0x3fffffff;

    for(int i=;i<n-;i++)

    {

        for(int j=i+;j<n;j++)

        {

            for(int k=;k<=m;k++)

            {

              dp[k]=a[j][k-]-a[i][k-];

              dp[k]=dp[k]+dp[k-];

              if(dp[k]>ans) ans=dp[k];

              if(dp[k]<) dp[k]=;

            }

        }

    }

    printf("%d\n",ans);

    return ;

}

令a[i][k]保存第k列，前i行的和。这样将矩阵通过a[j][k]-a[i][k]压缩成一维，然后求最大字串和。。

巴特西

最大子矩阵 hdu1081

To The Max

最新文章

热门文章