可以找到所有的解决方案,而无需探索搜索树中所有不成功的路径。下面是C++代码,虽然不高度优化,但几乎瞬间在我的电脑上找到了2个解决方案(结果是相同的唯一解决方案,因为有一个重复的拼图块,正确吗?)。
这里避免探索所有可能性的技巧是在我们放置拼图块时调用函数isValidSolution()(该函数处理空拼图块)。此外,为了加速进程,我按照给定顺序放置拼图块,从中间开始,然后在左侧、右侧、顶部和底部形成交叉点,然后是左上角、右上角、左下角和右下角的角落。 可能其他组合可以更快地执行。
当然,由于这个难题中的特殊模式分布(只接受一个可能的匹配的字母模式),可以进行优化,但这超出了我的回答范围。
#include<iostream>
#define PINK 1
#define YELLOW 2
#define BLUE 3
#define GREEN 4
#define LACOSTE 5
typedef int8_t pattern_t;
typedef struct {
pattern_t p[4];
} piece_t;
unsigned long long int solutionsCounter = 0;
piece_t emptyPiece = {.p = {0, 0, 0, 0} };
piece_t board[3][3] = {
{ emptyPiece, emptyPiece, emptyPiece},
{ emptyPiece, emptyPiece, emptyPiece},
{ emptyPiece, emptyPiece, emptyPiece},
};
inline bool isEmpty(const piece_t& piece) {
bool result = (piece.p[0] == 0);
return result;
}
bool isValidSolution() {
int i, j;
for (i = 0; i < 2; i++) {
for (j = 0; j < 3; j++) {
if (!isEmpty(board[i][j]) && !isEmpty(board[i+1][j]) && (board[i][j].p[1] != -board[i+1][j].p[3])) {
return false;
}
}
}
for (i = 0; i < 3; i++) {
for (j = 0; j < 2; j++) {
if (!isEmpty(board[i][j]) && !isEmpty(board[i][j+1]) && (board[i][j].p[2] != -board[i][j+1].p[0])) {
return false;
}
}
}
return true;
}
void rotatePiece(piece_t& piece) {
pattern_t paux = piece.p[0];
piece.p[0] = piece.p[1];
piece.p[1] = piece.p[2];
piece.p[2] = piece.p[3];
piece.p[3] = paux;
}
void printSolution() {
printf("Solution:\n");
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
printf("\t %2i ", (int) board[j][i].p[0]);
}
printf("\n");
for (int j = 0; j < 3; j++) {
printf("\t%2i %2i", (int) board[j][i].p[3], (int) board[j][i].p[1]);
}
printf("\n");
for (int j = 0; j < 3; j++) {
printf("\t %2i ", (int) board[j][i].p[2]);
}
printf("\n");
}
printf("\n");
}
bool usedPiece[9] = { false, false, false, false, false, false, false, false, false };
int colocationOrder[9] = { 4, 3, 5, 1, 7, 0, 2, 6, 8 };
void putNextPiece(piece_t pieces[9], int pieceNumber) {
if (pieceNumber == 9) {
if (isValidSolution()) {
solutionsCounter++;
printSolution();
}
} else {
int nextPosition = colocationOrder[pieceNumber];
int maxRotations = (pieceNumber == 0) ? 1 : 4;
for (int pieceIndex = 0; pieceIndex < 9; pieceIndex++) {
if (!usedPiece[pieceIndex]) {
usedPiece[pieceIndex] = true;
for (int rotationIndex = 0; rotationIndex < maxRotations; rotationIndex++) {
((piece_t*) board)[nextPosition] = pieces[pieceIndex];
if (isValidSolution()) {
putNextPiece(pieces, pieceNumber + 1);
}
rotatePiece(pieces[pieceIndex]);
}
usedPiece[pieceIndex] = false;
((piece_t*) board)[nextPosition] = emptyPiece;
}
}
}
}
int main() {
piece_t pieces[9] = {
{.p = { -YELLOW, -BLUE, +GREEN, +PINK} },
{.p = { -YELLOW, -GREEN, +PINK, +BLUE} },
{.p = { -BLUE, -YELLOW, +PINK, +GREEN }},
{.p = { -GREEN, -BLUE, +PINK, +YELLOW }},
{.p = { -PINK, -LACOSTE, +GREEN, +BLUE }},
{.p = { -PINK, -BLUE, +GREEN, +LACOSTE }},
{.p = { -PINK, -BLUE, +PINK, +YELLOW }},
{.p = { -GREEN, -YELLOW, +GREEN, +BLUE }},
{.p = { -GREEN, -BLUE, +PINK, +YELLOW }}
};
putNextPiece(pieces, 0);
printf("found %llu solutions\n", solutionsCounter);
return 0;
}
int availabilityPerPiece[]
而不是bool usedPiece[]
来处理重复的瓦片,并找到唯一的解决方案。 - Iban Cereijo