HDU 1081(动态规划-最大和子矩阵)

秒速五厘米 2022-06-02 07:07 160阅读 0赞

问题描述:

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

题目题意:让我们求出子矩阵中的最大和。

题目分析:我们枚举出每一个子矩阵，把每一个子矩阵的每一行的和算出来，我们就得到了一个m行的一维数组，我们求的就是这个序列的最大连续子序列，这个值就是这个子矩阵的最大和。

代码如下:

#include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    
    const int maxn=150;
    int a[maxn][maxn],dp[maxn],sum[maxn][maxn];
    
    int solve(int n)
    {
        int sum=0,ans=-0x3f3f3f3f;
        for (int i=1;i<=n;i++) {
           if (sum>=0) sum+=dp[i];
           else sum=dp[i];
          ans=max(ans,sum);
        }
        return ans;
    }
    int main()
    {
        int n;
        while (scanf("%d",&n)!=EOF) {
            memset (sum,0,sizeof (sum));
            for (int i=1;i<=n;i++) {
                for (int j=1;j<=n;j++){
                    scanf("%d",&a[i][j]);
                    sum[i][j]=sum[i][j-1]+a[i][j];//算前缀和，方便累加
                }
            }
            int ans=-0x7fffffff;
            for (int i=1;i<=n;i++) {//枚举行从i到n
                for (int j=1;j<=n;j++) {//从k到j
                    for (int k=1;k<=j;k++) {
                        int cnt=0;
                        for (int o=i;o<=n;o++) {
                            dp[++cnt]=sum[o][j]-sum[o][k-1];
                        }
                        ans=max(ans,solve(cnt));
                    }
                }
            }
            printf("%d\n",ans);
        }
        return 0;
    }