N-Queens II

N-Queens II


Follow up for N-Queens problem.

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

N-QueensII

N皇后问题,回溯可解。

AC代码:

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int getState(int n, int* board, int r, int c)
{

return board[r * n + c];
}

void setState(int n, int* board, int r, int c, int state)
{

board[r * n + c] = state;
}

void printBoard(int* board, int n)
{

printf("***************\n\n");
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
{
printf("%d ", getState(n, board, i, j));
}
printf("\n");
}
}

bool palced(int n, int* board, int r, int c)
{

for (int i = 0; i < n; ++i)
{
if (getState(n, board, r, i))
return false;
}
for (int i = 0; i < n; ++i)
{
if (getState(n, board, i, c))
return false;
}
int ir = r + 1;
int ic = c + 1;
// 斜线为0
while (ir < n && ic < n)
{
if (getState(n, board, ir, ic))
return false;
++ir;
++ic;
}
ir = r - 1;
ic = c - 1;
while (ir >= 0 && ic >= 0)
{
if (getState(n, board, ir, ic))
return false;
--ir;
--ic;
}
ir = r - 1;
ic = c + 1;
while (ir >= 0 && ic < n)
{
if (getState(n, board, ir, ic))
return false;
--ir;
++ic;
}
ir = r + 1;
ic = c - 1;
while (ir < n && ic >= 0)
{
if (getState(n, board, ir, ic))
return false;
++ir;
--ic;
}
}

void queens(int n, int* board, int* nums, int r, int c)
{

if ((r >= 0 && r < n) && ((c >= 0 && c < n)))
{
if (r == n - 1)
{
for (int i = 0; i < n; ++i)
{
if (palced(n, board, r, i))
{
//setState(n, board, r, i, 1);
++(* nums);
//printBoard(board, n); //可以打印棋盘
//setState(n, board, r, i, 0);
}
}
}
else
{
for (int i = 0; i < n; ++i)
{
if (palced(n,board,r,i))
{
setState(n, board, r, i, 1);//放置一个皇后
queens(n, board, nums, r + 1, 0);//对剩下的棋盘进行摆放
setState(n, board, r, i, 0);//恢复棋盘
}
}
}
}
}

int totalNQueens(int n) {
if (n ==0)
{
return 1;
}
int* board = (int*)malloc(sizeof(int) * n * n);
for (int i = 0; i < n*n; ++i)
{
board[i] = 0;
}
int result = 0;

queens(n, board, &result, 0, 0);

return result;
}