时间轴

2025-11-30

init


题目:

经典回溯问题,考虑每一行只能且必须放一个皇后,然后每次回溯考虑下一行放哪里。
注意标记能被攻击的区域时要-1而不是赋值为-1。这是避免两个皇后可以同时攻击一个区域。

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#include <vector>
using std::vector;

class Solution {
private:
int cnt;
void backtrace(vector<vector<int> > &board, int row, int n)
{
if (row == n) {
this->cnt++;
return;
}
int i, j, k;
for (j = 0; j < n; j++) { // column
if (board[row][j] == 0) {
for (i = row + 1; i < n; i++) { //set column of other row
board[i][j] -= 1;
}
i = row + 1;
k = j + 1;
while (i < n && k < n) {
board[i][k] -= 1;
i++;
k++;
}
i = row + 1;
k = j - 1;
while (i < n && k >= 0) {
board[i][k] -= 1;
i++;
k--;
}

board[row][j] = 1;

backtrace(board, row + 1, n);

i = row + 1;
k = j + 1;
while (i < n && k < n) {
board[i][k] += 1;
i++;
k++;
}
i = row + 1;
k = j - 1;
while (i < n && k >= 0) {
board[i][k] += 1;
i++;
k--;
}
for (i = row + 1; i < n; i++) { //set column of other row
board[i][j] += 1;
}
board[row][j] = 0;
}
}
}

public:
int totalNQueens(int n)
{
this->cnt = 0;

vector<vector<int> > board = vector(n, vector<int>(n, 0));
backtrace(board, 0, n);
return cnt;
}
};

int main()
{
Solution S;
S.totalNQueens(4);
}