To The Max

迷南。 2022-06-12 02:47 103阅读 0赞

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 −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

题意：找最大的矩阵和

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int n;
    int maxsum(int f[])
    {
        int dp[101],sum=f[1];
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            dp[i]=max(dp[i],dp[i-1]+f[i]);
            sum=max(sum,dp[i]);
        }
        return sum;
    }
    int main()
    {
        int a[101][101];
        while(~scanf("%d",&n))
        {
            memset(a,0,sizeof(a));
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    scanf("%d",&a[i][j]);
                    a[i][j]=a[i-1][j]+a[i][j];//将前面的每一行累加的下一行
                }
            }
            int s[101],maxx=a[1][1];
            for(int i=1;i<=n;i++)//遍历每一个区域
            {
                memset(s,0,sizeof(s));
                for(int j=i;j<=n;j++)
                {
                    for(int l=1;l<n;l++)
                        s[l]=a[j][l]-a[i-1][l];
                    maxx=max(maxx,maxsum(s));
                }
            }
            printf("%d\n",maxx);
        }
        return 0;
    }