UVa 10827 Maximum sum on a torus (贪心&矩阵环中的最大子矩阵和)-优快云博客

本文链接：https://blog.youkuaiyun.com/synapse7/article/details/11852177

本文介绍了一种解决矩形环中寻找最大和子矩形的问题，通过将矩形环扩展为2N*2N的矩阵并进行特殊处理来避免超出N*N的限制，最终实现了高效求解。

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10827 - Maximum sum on a torus

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1768

A grid that wraps both horizontally and vertically is called a torus. Given a torus where each cell contains an integer, determine the sub-rectangle with the largest sum. The sum of a sub-rectangle is the sum of all the elements in that rectangle. The grid below shows a torus where the maximum sub-rectangle has been shaded.

1	-1	0	0	-4
2	3	-2	-3	2
4	1	-1	5	0
3	-2	1	-3	2
-3	2	4	1	-4

Input

The first line in the input contains the number of test cases (at most 18). Each case starts with an integer N (1≤N≤75) specifying the size of the torus (always square). Then follows N lines describing the torus, each line containing N integers between -100 and 100, inclusive.

Output

For each test case, output a line containing a single integer: the maximum sum of a sub-rectangle within the torus.

Sample Input Output for Sample Input

1 -1 0 0 -4

2 3 -2 -3 2

4 1 -1 5 0

3 -2 1 -3 2

-3 2 4 1 -4

1 2 3

4 5 6

7 8 9

UVa 108 的加强版。

对于矩阵环，首先要预处理一下：把矩阵变成2N*2N。

并且要注意不要让子矩阵超过N*N。

完整代码：

/*0.079s*/

#include<cstdio>
#include<algorithm>
using namespace std;

int arr[155][155];

int main(void)
{
	int t, n, nn, maxn, sum;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		nn = n << 1;
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j)
			{
				scanf("%d", &arr[i][j]);
				arr[n + i][j] = arr[i][j]; ///列复制
				arr[i][j] += arr[i - 1][j]; ///列叠加
				arr[i][n + j] = arr[i][j]; ///行复制
			}
		for (int i = n + 1; i <= nn; ++i)
			for (int j = 1; j <= n; ++j)
			{
				arr[i][j] += arr[i - 1][j]; ///接上，继续列叠加
				arr[i][n + j] = arr[i][j]; ///行复制
			}
		maxn = 0;
		for (int i = 1; i <= n; i++)
			for (int j = i; j < n + i; j++)
			{
				sum = 0;
				for (int beg = 1, k = 1; beg <= n && k < nn; k++)
				{
					if (k - beg >= n)///子矩阵不要爆表，爆了的话就从beg标记处开始
					{
						k = ++beg;
						sum = 0;
					}
					sum += arr[j][k] - arr[i - 1][k];
					sum = max(0, sum);
					if (sum == 0) beg = k + 1;
					maxn = max(maxn, sum);
				}
			}
		printf("%d\n", maxn);
	}
	return 0;
}