ZOJ 1074 To the Max（最大子矩阵和）【java】

Problem

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.  

The input consists of an N x 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.

Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 

题意：给你一个矩阵，让你求最大子矩阵是多少

思路：枚举任意两列， 对这两列里的1-n行做最大子段和

import java.util.Scanner;

class Main{
	static final int maxn = 205;
	static final int inf = 10000000;
	static int a[][] = new int[maxn][maxn];
	static int sum[] = new int[maxn];
	public static void main(String[] args) {
		int n;
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()) {
			n = sc.nextInt();
			for(int i =  1; i <= n; i++)
				for(int j = 1; j <= n; j++)
					a[i][j] = sc.nextInt();
			int ans = -inf;
			for(int i = 1; i <= n; i++) {
				
				for(int j = i; j <= n; j++) {
					int cur = 0;
					for(int k = 1; k <= n; k++) {
						sum[k] = i == j ? a[k][i] : sum[k] + a[k][j];
						cur += sum[k];
						if(ans < cur) ans = cur;
						if(cur < 0) cur = 0;
					}
				}
			}
			System.out.println(ans);
		}
	}
}