我正在尝试找到数独的可能解决方案，它总是告诉我：0。
我必须用回溯来做，我不知道问题出在哪里，这是代码：
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class PracticaDAA3 {
public static boolean Cumple(int[][] matriz, int numero, int[] posicion) {
int x = posicion[0];
int y = posicion[1];
for (int i = 0; i < 9; i++) {
if (matriz[i][y] == numero) {
return false;
}
}
for (int j = 0; j < 9; j++) {
if (matriz[x][j] == numero) {
return false;
}
}
if (x < 3 && y < 3) {
for (int k = 0; k < 3; k++) {
for (int l = 0; l < 3; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x > 2 && x < 6 && y < 3) {
for (int k = 3; k < 6; k++) {
for (int l = 0; l < 3; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x > 5 && y < 3) {
for (int k = 6; k < 9; k++) {
for (int l = 0; l < 3; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x < 3 && y > 2 && y < 6) {
for (int k = 0; k < 3; k++) {
for (int l = 3; l < 6; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x > 2 && x < 6 && y > 2 && y < 6) {
for (int k = 3; k < 6; k++) {
for (int l = 3; l < 6; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x > 5 && y > 2 && y < 6) {
for (int k = 6; k < 9; k++) {
for (int l = 3; l < 6; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x < 3 && y > 5) {
for (int k = 0; k < 3; k++) {
for (int l = 6; l < 9; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x > 2 && x < 6 && y > 5) {
for (int k = 3; k < 6; k++) {
for (int l = 6; l < 9; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
} else if (x > 5 && y > 5) {
for (int k = 6; k < 9; k++) {
for (int l = 6; l < 9; l++) {
if (matriz[k][l] == numero) {
return false;
}
}
}
}
return true;
}
public static void siguientePosicion(int[] xy) {
int x = xy[0];
int y = xy[1];
//if(x<9 && y<9){
if (y == 8) {
xy[1] = 0;
xy[0]++;
//System.out.println("xy0: "+xy[0]+" xy1: "+xy[1]);
} else
xy[1]++;
}
public static void ResolSudoku(int[] xy, int[][] matriz, int[] solucion) {
int x = xy[0];
int y = xy[1];
if (x < 9 && y < 9) {
if (matriz[x][y] == 0) { //entramos por aqui si hay un 0
for (int i = 1; i < 10; i++) {
if (Cumple(matriz, i, xy)) { //aqui compruebo si mifila o columna o matriz tiene un nº
matriz[x][y] = i;
//System.out.println("xy0: "+xy[0]+" xy1: "+xy[1]);
if (x == 8 && y == 8)
solucion[0]++;
else {
int[] xy2 = new int[2];
xy2[0] = x;
xy2[1] = y;
siguientePosicion(xy2);
ResolSudoku(xy2, matriz, solucion);
}
matriz[x][y] = 0;
}
}
} else {
siguientePosicion(xy);
// System.out.println("X: "+x);
//System.out.println("x,y: "+x+y);
ResolSudoku(xy, matriz, solucion);
}
}
}
/**
* @param args the command line arguments
* @throws java.io.IOException
*/
public static void Imprimir(int[][] m) {
for (int i = 0; i < 9; i++) {
System.out.println();
for (int j = 0; j < 9; j++) {
System.out.print(m[i][j] + " ");
}
}
}
public static void main(String[] args) throws IOException {
int[][] matriz = new int[9][9];
int[] xy = new int[2];
xy[0] = 0;
xy[1] = 0;
BufferedReader entrada = new BufferedReader(new InputStreamReader(System.in));
for (int i = 0; i < 9; ++i) {
String[] lista = entrada.readLine().split(" ");
for (int j = 0; j < 9; j++) {
matriz[i][j] = Integer.parseInt(lista[j]);
}
}
int[] solucion = new int[1];
solucion[0] = 0;
ResolSudoku(xy, matriz, solucion);
System.out.println(solucion[0]);
Imprimir(matriz);
}
}
我知道这个输入：
5 3 4 0 7 0 0 0 0
6 0 0 1 9 5 0 0 0
0 9 8 0 0 0 0 6 0
8 0 0 0 6 0 0 0 3
4 0 0 8 0 3 0 0 1
7 0 0 0 2 0 0 0 6
0 6 0 0 0 0 2 8 0
0 0 0 4 1 9 0 0 5
0 0 0 0 8 0 0 7 9
输出必须为1，但结果为0。
有什么想法吗？
答案 0 :(得分：0)
问题是，如果matriz[x][y]中的数字为零，则只检查您是否有有效的解决方案。在您的示例谜题中，matriz[8][8]为9，因此当if (x == 8 && y == 8) {和x都是y时，您的代码永远不会到达8行。
基本上，您有这个if语句（代码替换为注释）：
if (matriz[x][y] == 0) {
// Check to see what numbers fit here. For each number that fits here, check
// to see if we're at the last grid cell. If so, record a solution, otherwise
// move to the next grid cell.
}
else {
// Move to the next grid cell.
}
修复方法是用以下内容替换else子句：
else {
// Check to see if we're at the last grid cell. If so, record a solution,
// otherwise move to the next grid cell.
}