[LeetCode] N-Queens II

N-Queens II

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

solution:

#include <iostream>

#include <iomanip>

#include <list>

using namespace std;

 

//board is 0 for avaliable positons, 1 for the other

int board[100][100];

int queenPos[100];

int N;

int solutionNum = 0;

 

void dfs(int k)

{

if(k == N)

{//the last queen has been put properly

solutionNum++;

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;

}

 

}

 

//return the total number of distinct solutions for N-Queens problem.

int totalNQueens(int n) {

N = n;

solutionNum = 0;

for(int i = 0;i < 100;i++)

for(int j = 0;j < 100;j++)

board[i][j] = 0;

 

dfs(0);

return solutionNum;        

}

 

int main()

{

for(int i = 0;i < 100;i++)

memset(board[i], 0, sizeof(int) * 100);

memset(queenPos, 0, sizeof(int) * 100);

for(int i = 1;i < 14;i++)

{

totalNQueens(i);

cout << "solutionNum = " << solutionNum << endl << endl;

}

return 0;

}