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2013
11-11

POJ 2260 Error Correction [解题报告] Java

Error Correction

问题描述 :

A boolean matrix has the parity property when each row and each column has an even sum, i.e. contains an even number of bits which are set. Here’s a 4 x 4 matrix which has the parity property:
1 0 1 0

0 0 0 0
1 1 1 1
0 1 0 1

The sums of the rows are 2, 0, 4 and 2. The sums of the columns are 2, 2, 2 and 2.

Your job is to write a program that reads in a matrix and checks if it has the parity property. If not, your program should check if the parity property can be established by changing only one bit. If this is not possible either, the matrix should be classified as corrupt.

输入:

The input will contain one or more test cases. The first line of each test case contains one integer n (n<100), representing the size of the matrix. On the next n lines, there will be n integers per line. No other integers than 0 and 1 will occur in the matrix. Input will be terminated by a value of 0 for n.

输出:

For each matrix in the input file, print one line. If the matrix already has the parity property, print “OK”. If the parity property can be established by changing one bit, print “Change bit (i,j)” where i is the row and j the column of the bit to be changed. Otherwise, print “Corrupt”.

样例输入:

4
1 0 1 0
0 0 0 0
1 1 1 1
0 1 0 1
4
1 0 1 0
0 0 1 0
1 1 1 1
0 1 0 1
4
1 0 1 0
0 1 1 0
1 1 1 1
0 1 0 1
0

样例输出:

OK
Change bit (2,3)
Corrupt

解题代码:

//* @author 洪晓鹏<[email protected]>
import java.util.Scanner;

public class Main {

/**
 * @param args
 */
public static void main(String[] args) {
 // TODO Auto-generated method stub
	Scanner in = new Scanner(System.in);
	while (true) {
		int n = in.nextInt();
		if (n == 0)
			break;
		Byte[][] matrix = new Byte[n][n];
		int row = 0;
		int sign1 = 0;
		int column = 0;
		int sign2 = 0;
		for (int i = 0; i < n; i++) {
			int sum = 0;
			for (int j = 0; j < n; j++) {
			  matrix[i][j] = in.nextByte();
			  sum += matrix[i][j];
			}
			if (sum % 2 == 1) {
				sign1 = i + 1;
				row++;
			}
		}
		if (row > 1) {
		   System.out.println("Corrupt");
		   continue;
		}
		for (int i = 0; i < n; i++) {
			int sum = 0;
			for (int j = 0; j < n; j++) {
               			sum += matrix[j][i];
			}
			if (sum % 2 == 1) {
				sign2 = i + 1;
				column++;
			}
		}
		if (column > 1) {
			System.out.println("Corrupt");
			continue;
		}
		if (column == 1 && row == 1)
		 System.out.println("Change bit (" + sign1 + "," + sign2 + ")");
		if ((column | row) == 0)
			System.out.println("OK");
		}
	}

}

  1. 这道题目的核心一句话是:取还是不取。
    如果当前取,则index+1作为参数。如果当前不取,则任用index作为参数。