[LeetCode] N-Queens

N-Queens

 

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The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,

There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

 

Solution:

//board is 0 for avaliable positons, 1 for the other
int board[100][100];
int queenPos[100];
int N;
int solutionNum = 0;
vector<vector<string> > result;


void dfs(int k)
{
    if(k == N)
    {//the last queen has been put properly
        solutionNum++;
        vector<string> cur;
        for(int i = 0;i < N;i++)
        {
            string column= "";
            for(int j = 0;j < N;j++)
            {
                if(queenPos[i] == j)
                    column += "Q";
                else
                    column += ".";
            }
    //        cout << column << endl;
            cur.push_back(column);
        }
    //    cout << endl << endl;
        result.push_back(cur);
        return;
    }

    for(int i = 0;i < N;i++)
    {
        //search for a postion for the current queen
        bool flag = true;

        if(board[i][k] == 0)
        {
            list<int> lastX, lastY;
            //this is a good postion for the k th column queen.
            queenPos[k] = i;//in the ith row, kth column
            //change the board state to mark its attark region.
            for(int j = 0;j < N;j++)
            {
                if(board[i][j] == 0)
                {
                    board[i][j] = 1;
                    lastX.push_back(i);
                    lastY.push_back(j);
                //    cout << i << " " << j << endl;
                }
                if(board[j][k] == 0)
                {
                    board[j][k] = 1;
                    lastX.push_back(j);
                    lastY.push_back(k);
                    //cout << j << " " << k << endl;
                }
                //
                if(i - j >= 0 && k - j >= 0 && board[i - j][k - j] == 0)
                {
                    //left up
                    lastX.push_back(i - j);
                    lastY.push_back(k - j);
                    board[i - j][k - j] = 1;

                    //cout << i -j  << " " << k - j << endl;
                }
                if(i - j >= 0 && k + j < N && board[i - j][k + j] == 0)
                {
                    lastX.push_back(i - j);
                    lastY.push_back(k + j);
                    board[i - j][k + j] = 1;
                    //cout << i -j  << " " << k + j << endl;
                }
                if(i + j < N && k - j >= 0 && board[i + j][k - j] == 0)
                {
                    lastX.push_back(i + j);
                    lastY.push_back(k - j);
                    board[i + j][k - j] = 1;
                //    cout << i + j  << " " << k - j << endl;
                }
                if(i + j < N && k + j < N && board[i + j][k + j] == 0)
                {
                    lastX.push_back(i + j);
                    lastY.push_back(k + j);
                    board[i + j][k + j] = 1;
                //    cout << i + j  << " " << k + j << endl;
                }
            }
            
            //cout << "put the " << k << " queen at " << i << " size = " << lastX.size() << endl;

            dfs(k + 1);
            
            //cout << "size = " << lastX.size() << endl;
            //back to the previous state.
            int num = lastX.size();
            for(int t = 0;t < num;t++)
            {
                int x = lastX.front();
                lastX.pop_front();
                int y = lastY.front();
                lastY.pop_front();
                board[x][y] = 0;
        //        cout << x << " " << y << endl;
            }    
        }
        else
            continue;
    }

}

vector<vector<string> > solveNQueens(int n) {
        N = n;
    solutionNum = 0;
    for(int i = 0;i < 100;i++)
    {
        for(int j = 0;j < 100;j++)
            board[i][j] = 0;
        queenPos[i] = 0;
    }
    dfs(0);
    return result;    
    }
View Code

 

posted @ 2014-07-05 07:45  xchangcheng  阅读(198)  评论(0编辑  收藏  举报