I solved the N- Queen problem with the condition that there can only be one queen per column. So I place a queen in a square in first column, then move onto the next column and place a queen in a square not attacked by the queen on board. I am able to find all solutions with this approach but it starts taking a long tim开发者_JAVA技巧e after n=13. Also I found that most of the solutions of the problem can be found by rotations and reflections of a very few distinct solutions.E.g 8 queen problem has 92 total solutions out of which only 12 are distinct. (http://en.wikipedia.org/wiki/Eight_queens_puzzle)
So my question is how do I check for these states of the board and only push those states onto the stack which give a distinct solution?
This is what I am doing right now.
typedef struct state{
int board[N][N];
int col;
}state;
state cur;
state next;
stack<state> myS;
myS.push(emptyBoard);
while(!myS.empty()){
cur=myS.top(); myS.pop();
if(cur.col==n){
count++;
continue;
}
for(i=0;i<n;i++){
next=cur;
if(cur.board[i][cur.col]==available){
next.board[i][cur.col]=occupied;
markConflicts(i,cur.col); //mark squares attacked by queen as conflicted
next.col=cur.col+1;
myS.push(next);
}
}
}
Well, you can only check for a unique solution once you have a solution. The place to check is at the point you increment the count
variable. At this point, compare the current board to a set of unique boards and if it's not in the set then add the new solution to it.
As for your speed, your solution has a bottleneck when pushing and popping the state
value. The bigger the board, the slower this becomes.
A much faster way is to only have one board and have each square keep a count og the number of queens controlling the square. So you'd have this:
function scan (column)
if column == number of columns (zero based index)
check solution
else
if there's a free space
add queen
increment controlled squares in columns to right of current column
scan (column+1)
decrement controlled squares in columns to right of current column
remove queen
This has far less data being pushed / popped and will greatly increase the speed.
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