# POJ 1050 To the Max C++版

POJ题解 4062浏览

DP，最大子段和，最大子矩阵和

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:

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

Slyar:给出n*n的矩阵，求和最大的子矩阵。

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]}

2、考虑二维的最大子矩阵问题

```#include
#define SIZE 101

using namespace std;

int a[SIZE][SIZE];

/* 最大子段和 */
int MaxSubArray(int n, int* a)
{
int max = 0;
int b = 0;
for (int i = 0; i < n; i++)
{
if (b > 0) b += a[i];
else b = a[i];
if (b > max) max = b;
}
return max;
}

/* 最大子矩阵和 */
int MaxSubMatrix(int m, int n)
{
int i, j, k, sum;
int max = 0;
int b[SIZE];
/* i为起始行 */
for (i = 0; i < m; i++)
{
/* 起始行改变才初始化b[] */
for (k = 0; k < n; k++)
{
b[k] = 0;
}

/* 从第i行加到最后一行 */
for (j = i; j < m; j++)
{
/* 每次把第j行的数都加起来 */
for (k = 0; k < n; k++)
{
b[k] += a[j][k];
}

/* 求出加和的最大子段和 */
sum = MaxSubArray(k, b);
if (sum > max)
{
max = sum;
}
}
}
return max;
}

int main()
{
int n;
cin >> n;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cin >> a[i][j];
}
}
cout << MaxSubMatrix(n, n) << endl;
//system("pause");
return 0;
}
```

### 网友最新评论 (3)

1. 是的，合肥的那位朋友所指出的问题确实存在，将max初始化的时候设成数组的第一个元素即可
59snest8年前 (2013-07-08)回复
2. 我发现你的程序貌似没考虑输出为负数的情况哈 输入4 -1 -2 -3 -4 输出 -1 但你的程序输出是0哦，改一下吧。^-^
匿名9年前 (2012-07-13)回复
3. 讲解的很好，谢谢
匿名11年前 (2010-04-11)回复