允许用户为数独求解程序输入数据?

时间:2013-05-02 13:47:16

标签: c++ sudoku solver

我目前正在尝试允许用户在Sudoku解算器中输入值,但是我一直收到错误。这是代码:

   #include <iostream> //
   #include <fstream>

   using namespace std;

   class SudokuBoard;
   void printB(SudokuBoard sb);

   typedef unsigned int uint;

   const uint MAXVAL = 9;
   const uint L = 9;
   const uint C = 9;
   const uint S = L * C;
   const uint ZONEL = 3;
   const uint ZONEC = 3;
   const uint ZONES = ZONEL * ZONEC;



   const uint lineElements[L][C] = {
   { 0,  1,  2,  3,  4,  5,  6,  7,  8},
   { 9, 10, 11, 12, 13, 14, 15, 16, 17},
   {18, 19, 20, 21, 22, 23, 24, 25, 26},
   {27, 28, 29, 30, 31, 32, 33, 34, 35},
   {36, 37, 38, 39, 40, 41, 42, 43, 44},
   {45, 46, 47, 48, 49, 50, 51, 52, 53},
   {54, 55, 56, 57, 58, 59, 60, 61, 62},
   {63, 64, 65, 66, 67, 68, 69, 70, 71},
   {72, 73, 74, 75, 76, 77, 78, 79, 80}
  };

   const uint columnElements[C][L] = {
{ 0,  9, 18, 27, 36, 45, 54, 63, 72},
{ 1, 10, 19, 28, 37, 46, 55, 64, 73},
{ 2, 11, 20, 29, 38, 47, 56, 65, 74},
{ 3, 12, 21, 30, 39, 48, 57, 66, 75},
{ 4, 13, 22, 31, 40, 49, 58, 67, 76},
{ 5, 14, 23, 32, 41, 50, 59, 68, 77},
{ 6, 15, 24, 33, 42, 51, 60, 69, 78},
{ 7, 16, 25, 34, 43, 52, 61, 70, 79},
 { 8, 17, 26, 35, 44, 53, 62, 71, 80}
};

   const uint zoneElements[S / ZONES][ZONES] = {
{ 0,  1,  2,  9, 10, 11, 18, 19, 20},
{ 3,  4,  5, 12, 13, 14, 21, 22, 23},
{ 6,  7,  8, 15, 16, 17, 24, 25, 26},
{27, 28, 29, 36, 37, 38, 45, 46, 47},
{30, 31, 32, 39, 40, 41, 48, 49, 50},
{33, 34, 35, 42, 43, 44, 51, 52, 53},
{54, 55, 56, 63, 64, 65, 72, 73, 74},
{57, 58, 59, 66, 67, 68, 75, 76, 77},
{60, 61, 62, 69, 70, 71, 78, 79, 80}
};

   class SudokuBoard {
   public:
SudokuBoard() :
    filledIn(0)
{
    for (uint i(0); i < S; ++i)
        table[i] = usedDigits[i] = 0;
}

virtual ~SudokuBoard() {
}

int const at(uint l, uint c) { // Returns the value at line l and row c
    if (isValidPos(l, c))
        return table[l * L + c];
    else
        return -1;
}

void set(uint l, uint c, uint val) { // Sets the cell at line l and row c to hold the value val
    if (isValidPos(l, c) && ((0 < val) && (val <= MAXVAL))) {
        if (table[l * C + c] == 0)
            ++filledIn;
        table[l * C + c] = val;
        for (uint i = 0; i < C; ++i) // Update lines
            usedDigits[lineElements[l][i]] |= 1<<val;
        for (uint i = 0; i < L; ++i) // Update columns
            usedDigits[columnElements[c][i]] |= 1<<val;
        int z = findZone(l * C + c);
        for (uint i = 0; i < ZONES; ++i) // Update columns
            usedDigits[zoneElements[z][i]] |= 1<<val;
    }
}

void solve() {
    try { // This is just a speed boost
        scanAndSet(); // Logic approach
        goBruteForce(); // Brute force approach
    } catch (int e) { // This is just a speed boost
    }
}

void scanAndSet() {
    int b;
    bool changed(true);
    while (changed) {
        changed = false;
        for (uint i(0); i < S; ++i)
            if (0 == table[i]) // Is there a digit already written?
                if ((b = bitcount(usedDigits[i])) == MAXVAL - 1) { // If there's only one digit I can place in this cell, do
                    int d(1); // Find the digit
                    while ((usedDigits[i] & 1<<d) > 0)
                        ++d;
                    set(i / C, i % C, d); // Fill it in
                    changed = true; // The board has been changed so this step must be rerun
                } else if (bitcount(usedDigits[i]) == MAXVAL)
                    throw 666; // Speed boost
    }
}

void goBruteForce() {
    int max(-1); // Find the cell with the _minimum_ number of posibilities (i.e. the one with the largest number of /used/ digits)
    for (uint i(0); i < S; ++i)
        if (table[i] == 0) // Is there a digit already written?
            if ((max == -1) || (bitcount(usedDigits[i]) > bitcount(usedDigits[max])))
                max = i;

    if (max != -1) {
        for (uint i(1); i <= MAXVAL; ++i) // Go through each possible digit
            if ((usedDigits[max] & 1<<i) == 0) { // If it can be placed in this cell, do
                SudokuBoard temp(*this); // Create a new board
                temp.set(max / C, max % C, i); // Complete the attempt
                temp.solve(); // Solve it
                if (temp.getFilledIn() == S) { // If the board was completely solved (i.e. the number of filled in cells is S)
                    for (uint j(0); j < S; ++j) // Copy the board into this one
                        set(j / C, j % C, temp.at(j / C, j % C));
                    return; // Break the recursive cascade
                }
            }
    }
}

uint getFilledIn() {
    return filledIn;
}

   private:
uint table[S];
uint usedDigits[S];
uint filledIn;

bool const inline isValidPos(int l, int c) {
    return ((0 <= l) && (l < (int)L) && (0 <= c) && (c < (int)C));
}

uint const inline findZone(uint off) {
    return ((off / C / ZONEL) * (C / ZONEC) + (off % C / ZONEC));
}

uint const inline bitcount(uint x) {
    uint count(0);
    for (; x; ++count, x &= (x - 1));
    return count;
}
};


   void printB(SudokuBoard sb) {
cout << " ***** Sudoku Solver By Diego Freitas ***** " << endl;
cout << "  |  -------------------------------  |" << endl;
for (uint i(0); i < S; ++i) {
    if (i % 3 == 0)
        cout << "  |";
    cout << "  " << sb.at(i / L, i % L);
    if (i % C == C - 1) {
        if (i / C % 3 == 2)
            cout << "  |" << endl << "  |  -------------------------------";
        cout << "  |" << endl;
    }
}
cout << endl;
}

   int main(int argc, char *argv[]) {
SudokuBoard sb;

ifstream fin("Sudoku.in");
int aux;
for (uint i(0); i < S; ++i) {
    fin >> aux;
    sb.set(i / L, i % L, aux);
}

fin.close();

printB(sb);
sb.solve();
printB(sb);

system ("PAUSE");
return 0;

} 

如果有人对代码进行了任何更改,那么我们将不胜感激。我目前在我自己的时间制作这个,我大致想知道它是如何工作的。

0 个答案:

没有答案