POJ 1050

From: http://www.slyar.com/blog/poj-1050-cpp.html

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]}

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。

#include <iostream> #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; }