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2013
11-09

POJ 1050 To the Max [解题报告] Java

To the Max

问题描述 :

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.

输入:

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].

输出:

Output the sum of the maximal sub-rectangle.

样例输入:

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

8  0 -2

样例输出:

15

解题代码:

import java.util.Scanner;

 public class Main {

     int[][] a;
     int l;
     int max = Integer.MIN_VALUE;
     int t;
     int t1;
     int t2[];

     public Main() {
         Scanner scan = new Scanner(System.in);
         l = scan.nextInt();
         a = new int[l][l];
         for (int i = 0; i < l; i++) {
             for (int j = 0; j < l; j++) {
                 a[i][j] = scan.nextInt();
             }
         }
         search();
         System.out.println(max);
     }

     public void search() {
         for (int i = 0; i < l; i++) {
             for (int j = 0; j < l; j++) {
                 t1 = 0;
                 t2 = new int[l];
                 for (int k = i; k < l; k++) {
                     t1 += a[k][j];
                     t = t1;
                     if (t > max) {
                         max = t;
                     }
                     for (int m = j + 1; m < l; m++) {
                         t2[m] += a[k][m];
                         t += t2[m];
                         if (t > max) {
                             max = t;
                         }
                     }
                 }
             }
         }
     }

     public static void main(String[] args) {
         new Main();
     }
}

  1. 换句话说,A[k/2-1]不可能大于两数组合并之后的第k小值,所以我们可以将其抛弃。
    应该是,不可能小于合并后的第K小值吧

  2. Often We don’t set up on weblogs, but I would like to condition that this established up really forced me individually to do this! considerably outstanding publish

  3. 第一句可以忽略不计了吧。从第二句开始分析,说明这个花色下的所有牌都会在其它里面出现,那么还剩下♠️和♦️。第三句,可以排除2和7,因为在两种花色里有。现在是第四句,因为♠️还剩下多个,只有是♦️B才能知道答案。