NC50959 To the Max
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:
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;
}
本文来自博客园,作者:空白菌,转载请注明原文链接:https://www.cnblogs.com/BlankYang/p/16578024.html