pku1050 最大子矩阵问题

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16586		Accepted: 8423

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

Source

Greater New York 2001

很容易看出，这个问题其实就是最大字段和问题的一个变形（把一维的变形为二维），既然可以把它看做一个变形，我们容易想到将它还原，也就是把一个矩阵压缩，压缩成一个一维的序列，这样问题就一目了然了。

以下是我的AC代码：

#include<iostream> #include<stdio.h> using namespace std; int a[100][100]; int main() { int n; while(cin>>n) { for(int i=0;i<n;i++) for(int j=0;j<n;j++) scanf("%d",&a[i][j]); int result=-999999999; for(int i=0;i<n;i++) for(int j=i;j<n;j++) { int b[100]={0}; for(int e=0;e<n;e++) for(int k=i;k<=j;k++) b[e]+=a[k][e]; int sum=0,Max=-999999999; for(int e=0;e<n;e++) { sum+=b[e]; if(sum>Max) Max=sum; else if(sum<0) sum=0; } if(Max>result) result=Max; } cout<<result<<endl; } return 0; }