UVa 108 Maximum Sum (贪心&最大子矩阵和)

108 - Maximum Sum

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=44

Background

A problem that is simple to solve in one dimension is often much more difficult to solve in more than one dimension. Consider satisfying a boolean expression in conjunctive normal form in which each conjunct consists
of exactly 3 disjuncts. This problem (3-SAT) is NP-complete. The problem 2-SAT is solved quite efficiently, however. In contrast, some problems belong to the same complexity class regardless of the dimensionality of the problem.

The Problem

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size

or
greater located within the whole array. As an example, the maximal sub-rectangle of the array:



is in the lower-left-hand corner:



and has the sum of 15.

Input and Output

The input consists of an

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

integers separated by white-space
(newlines and spaces). These

integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right,
then all numbers on 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].

The output is 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

UVa 507的升级版。

测试数据又有问题。。就算你不满足“A sub-rectangle is any contiguous sub-array of size

or
greater located within the whole array”(子矩阵不能为零矩阵，但由于测试数据太弱可以为零矩阵)也能AC。

正确的代码：

/*0.015s*/

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

int num[105][105];

int main(void)
{
	int n, temp, maxn;
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j)
			{
				scanf("%d", &num[i][j]);
				num[i][j] += num[i - 1][j]; ///累加每一列，越往下累加的数越多
			}
		maxn = num[1][1];
		///从上往下，从左往右
		for (int i = 1; i <= n; ++i)
			for (int j = i; j <= n; ++j)///j从i开始
			{
				temp = 0;
				for (int k = 1; k <= n; ++k)
				{
					if (temp >= 0) temp += num[j][k] - num[i - 1][k];
					else temp = num[j][k] - num[i - 1][k];
					maxn = max(maxn, temp);
				}
			}
		printf("%d\n", maxn);
	}
	return 0;
}