【发布时间】：2013-04-15 15:44:00
【问题描述】：
我是新手，五年多没有编写 C++ 代码了。我有一个关于数独求解器的课程。它（在大多数情况下）有效，除非它回溯（当它无法找到放入框中的单个可能值时）它要么一直回溯到开头，要么无法退出循环。帮助？代码如下
Address::Address(int row, int col) //This class is for the addresses of each block of of the 9x9 grid
{
this->row=row;
this->col=col;
}//finds the locations of each block
Puzzle::Puzzle(const char grid[][9])
{
}
void Puzzle::solve(char grid[][9])
{
stack <Address> locations; //creates stack of locations
for(int row=0; row<9;row++)
{
for (int col=0; col<9; col++)
{
if (grid[row][col]=='*') //checks for an empty block
{
Possibles possibles(grid,Address(row,col)); //creates an array of possible values for that specific block
int possvalue = possibles.GetNextPossible(0); //gets the next possible value from the array of possible values
while(possvalue == -1) //checks if there's no possible values for the block
{
if(locations.empty())
{
cout << "Puzzle Unsolvable" << endl;
return;
}
cout<<"PossValue before: "<< possvalue <<endl;
Address previouslocation = locations.top(); //stores the previous location from the stack
Possibles previouspossible(grid,previouslocation); //creates a new array of new possibles
previouspossible.array[grid[previouslocation.row][previouslocation.col]-'0'] = false;
grid[previouslocation.row][previouslocation.col] = '*'; //changes the previous location back to empty
possvalue = previouspossible.GetNextPossible(grid[previouslocation.row][previouslocation.col]-'0'); //gets the new possible value from new possibles
locations.pop();
cout<<"PossValue after: "<< possvalue<<endl;
cout<<"row: "<< row<<endl;
cout<<"col: "<< col<<endl;
cout<<"previouslocation.row: "<< previouslocation.row<<endl;
cout<<"previouslocation.col: "<< previouslocation.col<<endl;
cout<<"previousvalue" << grid[previouslocation.row][previouslocation.col] << endl;
cout<<"grid[previouslocation.row][previouslocation.col]: "<< grid[previouslocation.row][previouslocation.col]<<endl;
cout << locations.size() << endl;
cout<<endl;
row = previouslocation.row; //changes the row where we're trying to solve
col = previouslocation.col; //changes the column where we're trying to solve
}
grid[row][col] = (char) ( ((int)'0') + possvalue); //puts the "good" value on the grid
locations.push (Address(row,col)); // pushes the location of the box into the stack
/*for(int i = 0; i< 9; i++)
{
for(int j = 0; j< 9; j++)
cout << grid[i][j];
cout << endl;
}*/ // for row
cout << endl;
}
}
}
}//solves sudoku
Possibles::Possibles(char grid[][9], Address currentAddress) //makes an array for each block for the number of possibles
{
for (int i=0; i<10; i++)
{
array[i] = true; //setting a boolean array of size 9 to true
}
for (int row=0; row<9;row++)
{
array[grid[row][currentAddress.col]-'0'] = false; //checks for repeated values on the row
}
for(int col=0; col<9; col++)
{
array[grid[currentAddress.row][col]-'0'] = false; //checks for repeated values on the column
}
for(int row=currentAddress.row-(currentAddress.row%3); row < (currentAddress.row-(currentAddress.row%3))+3 ; row++)
{
for(int col=currentAddress.col-(currentAddress.col%3); col < (currentAddress.col-(currentAddress.col%3))+3 ; col++)
{
array[grid[row][col]-'0'] = false; //checks for repeated values in the box
}
}
}//this function returns an array of "possibles" which is a boolean array for every value from 1-9. This determines whether the value is a possibility for that block.
int Possibles::GetNextPossible(int nextpossible) //determines the next "possible" array
{
for(int i=nextpossible+1; i<10; i++)
{
if (array[i] == true)
{
return i; //if the boolean array returns true, then return the value of that boolean array.
}
}
return -1; // if there is no "true" in the boolean array, then returns "NULL"
}
谢谢你们的帮助，伙计们！任何意见表示赞赏。
【问题讨论】：
-
你可能会给它一些无法解决的数独问题吗？
-
为了防止无限循环，你需要一些条件来保证你会停止。例如，如果您根据预定义的顺序遍历所有解决方案，并且您的循环在顺序上是单调的（比如说总是增加），那么它必须终止。 （与递归类似，这是确保您触底的标准方法。）
-
@NicholasWilson 在没有递归的情况下进行回溯非常困难。