POJ 1050 To the Max【最大子矩阵】

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

思路

将一维的求最大子序列的和进行加强版，同样这里采取相对暴力的方法，时间复杂度为O（N^3）的方法进行枚举每个子矩阵。

源码

#include<stdio.h> #include<string.h> #define MIN -99999999 int main() { int i, j, k, f[101][101], a[101], n, mem[101]; while(scanf("%d", &n)!=EOF) { for(i=1; i<=n; i++) for(j=1; j<=n; j++) scanf("%d", &f[i][j]); int maxnum=MIN, change; memset(a, 0, sizeof(a)); for(i=1; i<=n; i++) { memset(mem, 0, sizeof(mem)); for(j=i; j<=n; j++) { for(k=1; k<=n; k++) { mem[k]+=f[j][k]; a[k]=mem[k]; } change=a[1]; for(k=2; k<=n; k++) { a[k]=a[k]>a[k]+a[k-1]?a[k]:a[k]+a[k-1]; change=a[k]>change?a[k]:change; } maxnum=maxnum>change?maxnum:change; } } printf("%d\n", maxnum); } }