HDU 1081 to the max 基础DP 好题

　　　　　　　　　　　　　　　　　　To The Max

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

 

 

 

 

 

就是给一个n*n的矩阵，求出它的最大子矩阵

 

我们都做过求一个序列的最大子序列吧。O(n)的复杂度。

 

这道题就是那道题转化一下。

 

暴力枚举第k1行到第k2行每一列各个元素的和sum[i]，然后对sum[i]进行一次最大子序列求和

 

然后不断更新答案

 

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=105;

int a[maxn][maxn];
int sum[maxn];
int dp[maxn];

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&a[i][j]);

        int ans=-1000000;

        for(int i=1;i<=n;i++)
        {
            memset(dp,0,sizeof(dp));
            memset(sum,0,sizeof(sum));
            for(int j=i;j<=n;j++)
            {
                for(int k=1;k<=n;k++)
                {
                    sum[k]+=a[j][k];
                }

                for(int k=1;k<=n;k++)
                {
                    dp[k]=max(dp[k-1],0)+sum[k];
                    if(dp[k]>ans)
                        ans=dp[k];
                }

            }
        }

        printf("%d
",ans);
    }

    return 0;
}