n皇后问题

#include<iostream>
#include<vector>
#include<string>
using namespace std;
void put_queen(int n,int x, int y, vector<vector<int>>& attack) {
  int dx[] = { -1,-1,-1,0, 1, 1, 1, 0 };
  int dy[] = { -1,0, 1, 1, 1, 0,-1,-1 };
  attack[x][y] = 1;
  for (int i = 1; i <n; i++)//从1延伸到n-1的距离
    for (int j = 0; j < 8; j++) {//从八个方向延伸
      int nx = x + i * dx[j];
      int ny = y + i * dy[j];
      if (nx >= 0 && nx < n && ny >= 0 && ny < n) {
        attack[nx][ny] = 1;
      } 
    }
}
//算法思想:定义一个二维数组attack，0表示可以放置皇后不会被攻击，1表示不可以放置皇后会被攻击。定义一个二维数组queen记录皇后放置的位置，Q代表放置，代表空格。
//从第0行开始尝试放置皇后，如果当前位置attack=0则放置，先备份attack，然后更新，尝试下一行皇后放置，如果尝试结束则恢复attack，恢复取消当前位置的放置，如果n行尝试结束，则记录解法。
void backtrack(int k,int n,vector<vector<int>> &attack, vector<string>queen, vector<vector<string>>& solve) {
  if (k == n) {
    solve.push_back(queen);
  }
  else {
    for (int i = 0; i < n; i++) {
      if (attack[k][i] == 0) {
        vector<vector<int>>tmp = attack;//备份attack
        queen[k][i] = 'Q';//放置皇后
        put_queen(n,k, i, attack);//更新attack
        backtrack(k + 1, n, attack, queen, solve);//尝试下一行
        attack = tmp;//恢复attack
        queen[k][i] = '.';//恢复皇后位置
      }
    }
  }
}
vector<vector<string>> solveNQueens(int n) {
  vector<vector<string>> solve;
  vector<vector<int>> attack(n, vector<int>(n, 0));
  vector<string>queen;
  for (int i = 0; i < n; i++) {
    queen.push_back("");
    queen[i].append(n, '.');
  }
  backtrack(0, n, attack, queen, solve);
  return solve;
}
int main() {
  vector<vector<string>> solve=solveNQueens(8);
  for (auto it : solve) {
    for (int i = 0; i < it.size(); i++) {
      cout << it[i] << " ";
    }
  }
}