NC50959 To the Max

题目链接

题目

题目描述

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.

输入描述:

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].

输出描述

Output the sum of the maximal sub-rectangle.

示例1

输入

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

8  0 -2

输出

15

题解

知识点：线性dp，枚举，前缀和。

最大子串和的变种。可以枚举矩阵行端点的组合，然后对列做最大子串和。

先用数组 \(lsum\) 得到每列的前缀和。然后设 \(dp[i]\) 为某次行组合考虑到 \(i\) 列（选 \(i\) 列）的最大子矩阵和。转移方程为：

\[dp[k] = \max (dp[k-1] + lsum[k][j] - lsum[k][i-1])
\]

时间复杂度 \(O(n^3)\)

空间复杂度 \(O(n^2)\)

代码

#include <bits/stdc++.h>

using namespace std;

int lsum[107][107], dp[107];

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    for (int i = 1;i <= n;i++)
        for (int j = 1, tmp;j <= n;j++)
            cin >> lsum[j][i], lsum[j][i] += lsum[j][i - 1];
    int ans = -2e9;
    for (int i = 1;i <= n;i++) {
        for (int j = i;j <= n;j++) {
            dp[0] = 0;
            for (int k = 1;k <= n;k++)
                dp[k] = max(dp[k - 1], 0) + lsum[k][j] - lsum[k][i - 1];
            for (int k = 1;k <= n;k++)
                ans = max(ans, dp[k]);
        }
    }
    cout << ans << '\n';
    return 0;
}