[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;



}

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