1 #include "000庫函數.h"
  2 
  3 
  4 //使用回溯法來計算
  5 //經典解法為回溯遞歸，一層一層的向下掃描，需要用到一個pos數組，
  6 //其中pos[i]表示第i行皇後的位置，初始化為 - 1，然後從第0開始遞歸，
  7 //每一行都一次遍歷各列，判斷如果在該位置放置皇後會不會有沖突，以此類推，
  8 //當到最後一行的皇後放好後，一種解法就生成了，將其存入結果res中，
  9 //然後再還會繼續完成搜索所有的情況，代碼如下：17ms
 10 class Solution {
 11 public:
 12     vector<vector<string
 
>> solveNQueens(int n) {        
 13         vector<vector<string>>res;
 14         vector<int>pos(n, -1);
 15         NQueue(res, pos, 0);
 16         return res;
 17     }
 18     
 19     void NQueue(vector<vector<string>>&res, vector<int>&pos, int
 
 t) {
 20         int n = pos.size();
 21         if (t == n) {//組合成功
 22             vector<string>v(n, string(n, ‘.‘));//這初始化絕逼了
 23             for (int i = 0; i < n; ++i)
 24                 v[i][pos[i]] = ‘Q‘;
 25             res.push_back(v);
 26         }
 27         else
 28             for
 
 (int k = 0; k < n; ++k)
 29                 if (Danger(pos, t, k)) {
 30                     pos[t] = k;
 31                     NQueue(res, pos, t + 1);
 32                     pos[t] = -1;//切記，關鍵點，回溯
 33                 }
 34     }
 35 
 36     bool Danger(vector<int>pos, int t, int k) {
 37         for (int i = 0; i < t; ++i)
 38             if (pos[i] == k || abs(t - i) == abs(pos[i] - k))
 39                 return false;
 40         return true;
 41     }
 42 
 43 };
 44 
 45 
 46 //通過使用排列進行判斷是否可行進行求解
 47 //但是太耗時了，還是用回溯法吧
 48 class Solution {
 49 public:
 50     vector<vector<string>> solveNQueens(int n) {
 51         vector<vector<string>>res;
 52         vector<int>nums;
 53         for (int i = 0; i < n; ++i)
 54             nums.push_back(i);
 55 
 56         if (Danger(nums)) {
 57             vector<string>v;
 58             for (int i = 0; i < n; ++i) {
 59                 string s = "";
 60                 for (int k = 0; k < nums[i]; ++k)
 61                     s += ‘.‘;
 62                 s += ‘Q‘;
 63                 for (int k = nums[i] + 1; k < n; ++k)
 64                     s += ‘.‘;
 65                 v.push_back(s);
 66             }
 67             res.push_back(v);
 68         }
 69         while (next_permutation(nums.begin(), nums.end())) {
 70             if (Danger(nums)) {
 71                 vector<string>v;
 72                 for (int i = 0; i < n; ++i) {
 73                     string s = "";
 74                     for (int k = 0; k < nums[i]; ++k)
 75                         s += ‘.‘;
 76                     s += ‘Q‘;
 77                     for (int k = nums[i] + 1; k < n; ++k)
 78                         s += ‘.‘;
 79                     v.push_back(s);
 80                 }
 81                 res.push_back(v);
 82             }
 83         }
 84         return res;
 85     }
 86                 
 87 
 88 
 89     bool Danger(vector<int>nums) {//用來判斷是否可行
 90         for (int i = 0; i < nums.size(); ++i) {
 91             for (int j = 0; j < nums.size(); ++j) {
 92                 if (j == i)continue;
 93                 if ((j + nums[j]) == (i + nums[i]) || (i - nums[i]) == (j - nums[j]))
 94                     return false;
 95             }
 96         }
 97         return true;
 98     }
 99                     
100 
101 };
102 
103 
104 void T051() {
105     Solution s;
106     vector<vector<string>>v;
107     int n;
108     n = 5;
109     v = s.solveNQueens(n);
110     for (auto &a : v) {
111         for (auto b : a)
112             cout << b << endl;
113         cout << "//////////////////////////" << endl;
114     }
115     
116 }

力扣算法題—051N皇後問題