To the Max

 

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48507		Accepted: 25662

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. 

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 the sum of the maximal sub-rectangle.

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

8  0 -2

#include <iostream>
#include <cstring>

using namespace std;

int mmap[101][101];      //存矩阵
int dp[101][101][101];   //dp[k][i][j]存第k行 第i~j列的和。 比如dp[3][1][5]表示从 "第3行第1列" 到 "第3行第5列" 的和。
int sum[101][101][101];  //sum[k][i][j]表示"dp[1][i][j]"到"dp[k][i][j]"的和。比如sun[4][2][5]表示前4行 所有第2~5列的和。  注意：所有的数据输出从下标1开始。

int n;
int main()
{
    while(cin>>n)
    {
        memset(dp,0,sizeof(dp));  //数组初始化为0
        memset(sum,0,sizeof(sum));

        //矩阵数据输入
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin>>mmap[i][j];
            }
        }

        //计算dp数组
        for(int k=1;k<=n;k++)     //k表示第k行
        {
            for(int j=1;j<=n;j++)     //因为从i~j列所以j放在i的外层
            {
                for(int i=1;i<=j;i++)
                {
                    //统计从i~j列的和
                    int sum=0;
                    for(int p=i;p<=j;p++)
                        sum+=mmap[k][p];
                    dp[k][i][j]=sum;
                }

            }
        }

        //计算sum数组
        for(int j=1;j<=n;j++)
        {
            for(int i=1;i<=j;i++)
            {
                for(int k=1;k<=n;k++)
                {
                    //对应每组 i~j 列,前k行 所有的i~j列的元素的和
                    sum[k][i][j]+=(dp[k][i][j]+sum[k-1][i][j]);
                }
            }
        }

        int mmax=-1000000;

        for(int j=1;j<=n;j++)
        {
            for(int i=1;i<=j;i++)
            {
                //对应每组i~j列，统计每组 a~b 行的最大的和。
                for(int b=1;b<=n;b++)
                {
                    for(int a=1;a<=b;a++)
                    {
                        mmax=max(mmax,sum[b][i][j]-sum[a-1][i][j]);
                    }
                }
            }
        }

        cout<<mmax<<endl;

    }
    return 0;
}