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ACM学习-POJ-1050-To the Max
To the Max
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 37462 | Accepted: 19724 |
Description
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:
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].
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
Source
题目要求：求出最大子矩阵。
题目分析：
首先，如果遍历，找到所有的子矩阵，解出这道题是没问题的，只是在POJ上，必然会超时。
所以我考虑用动态规划来解决。即把这问题划分为更小的问题。
现在考虑把这个二维数组拆分，每行每行的先计算。
就把问题化解成求一维数组中子段和最大。 类似求最长字符串的问题。
然后累加比较总和。即可。
详细说明如下：
1、首先考虑一维的最大子段和问题，给出一个序列a[0]，a[1]，a[2]...a[n]，求出连续的一段，使其总和最大。
a[i]表示第i个元素
dp[i]表示以a[i]结尾的最大子段和
dp[i] = max{a[i], dp[i-1] + a[i]}
解释一下方程:
如果dp[i-1] > 0，则 dp[i] = dp[i-1] + a[i]
如果dp[i-1] < 0，则 dp[i] = a[i]
因为不用记录位置信息，所以dp[]可以用一个变量dp代替:
如果dp > 0，则dp += a[i]
如果dp < 0，则dp = a[i]
2、考虑二维的最大子矩阵问题
我们可以利用矩阵压缩把二维的问题转化为一维的最大子段和问题。因为是矩阵和，所以我们可以把这个矩形的高压缩成1，用加法就行了。
恩，其实这个需要自己画图理解，我的注释里写得很详细了，自己看吧。
枚举求的结果矩阵的行号范围，从(1,1), (1,2), ...到(1,n),然后再求(2,2),(2,3),...(2,n)的最大值，直至到(m,n)。
下面给出AC代码。
#include <stdio.h>
#define MAXSIZE 101
//求出一行中最大的子段和
int MaxArray( int n, int arr_[])
{
int i, sum_ = 0, max_ = 0;
for (i=1; i<=n; i++)
{
if (sum_>0)
{
sum_ += arr_[i];
}
else
{
sum_ = arr_[i];
}
if (sum_>max_)
{
max_ = sum_;
}
}
return max_;
}
//求出最大子矩阵和。
int MaxMatrix( int n, int arr_[][MAXSIZE])
{
int max_ = arr_[1][1];
int sum_;
int i, j, k;
int temp_arr[MAXSIZE];
for (i=1; i<=n; i++) //从第一行开始，直到第n行
{
for (j=1; j<=n; j++) //只有起始行改变，temp_arr数组才初始化
{
temp_arr[j] = 0;
}
for (j=i; j<=n; j++) //从i行到第n行
{
for (k=1; k<=n; k++)
{
temp_arr[k] += arr_[j][k]; //temp_arr[k] 表示从第i行到第n行中第k列的总和。
}
sum_ = MaxArray(n, temp_arr); //求出该行中最大的子段和
if (sum_ > max_)
{
max_ = sum_;
}
}
}
return max_;
}
int main()
{
int n;
int i, j;
int arr_[MAXSIZE][MAXSIZE];
int max_;
while (~scanf("%d", &n)) //多组测试。 相当于 scanf("%d", &n) != EOF
{
for (i=1; i<=n; i++)
{
for (j=1; j<=n; j++)
{
scanf("%d", &arr_[i][j]);
}
}
max_ = MaxMatrix(n, arr_);
printf("%d\n", max_);
}
return 0;
}