To the Max（dp）

最新推荐文章于 2024-07-31 10:00:00 发布

原创最新推荐文章于 2024-07-31 10:00:00 发布 · 183 阅读

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#dp

本文介绍了一种解决二维数组中寻找最大子矩阵和的问题的方法。通过对数组进行预处理，使用动态规划思想，枚举矩阵的长和宽，有效地找到了元素和最大的子矩阵。文章详细解释了算法思路，并提供了完整的C++代码实现。

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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.

Input
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
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

题意：在一个n*n的矩阵里找出他元素和最大的子矩阵。思路：老实说，这个题我一开始想到的是前缀和，然后枚举对角线两个点的坐标求最大值，但是我发现这样好像会t…… 正确的dp姿势应该是用dp数组记录第i行前j个数字的和，然后枚举矩阵的长和宽。长通过上下界确定，宽一步一步考虑，具体看代码。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<math.h>
#include<vector>
#include<stack>
#include<string>
#include<set>
#include<map>
#include<numeric>
#include<stdio.h>
#include<functional>
#include<time.h>
#pragma warning(disable:6031)

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1111;
const ll mode = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846264338327950;
template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template <class T> inline T min(T a, T b, T c, T d) { return min(min(a, b), min(c, d)); }
template <class T> inline T max(T a, T b, T c, T d) { return max(max(a, b), max(c, d)); }

int dp[150][150];
int main()
{
	int n, x;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
		{
			scanf("%d", &x);
			dp[i][j] = dp[i - 1][j] + x;
		}

	int ans = -1000;
	int sum = 0;
	for (int i = 1; i <= n; i++)
		for (int j = i; j <= n; j++)
		{
			sum = 0;
			for (int k = 1; k <= n; k++)
			{
				sum += dp[j][k] - dp[i - 1][k];//加上第k列的i到j的元素的和
				if (sum < 0) sum = 0;//如果小于零，这之前的就不用考虑了
				if (sum > ans) ans = sum;
			}
		}
	printf("%d\n", ans);


	return 0;
}