(PAT)N Queens Puzzle(N皇后改版)

傷城~ 2022-04-17 05:41 106阅读 0赞

The "eight queens puzzle" is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other. Thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example of the more general N queens problem of placing Nnon-attacking queens on an N×N chessboard. (From Wikipedia - "Eight queens puzzle".)

Here you are NOT asked to solve the puzzles. Instead, you are supposed to judge whether or not a given configuration of the chessboard is a solution. To simplify the representation of a chessboard, let us assume that no two queens will be placed in the same column. Then a configuration can be represented by a simple integer sequence (Q1,Q2,⋯,QN), where Qi is the row number of the queen in the i-th column. For example, Figure 1 can be represented by (4, 6, 8, 2, 7, 1, 3, 5) and it is indeed a solution to the 8 queens puzzle; while Figure 2 can be represented by (4, 6, 7, 2, 8, 1, 9, 5, 3) and is NOT a 9 queens' solution.

<table> 
 <thead> 
  <tr> 
   <th><img alt="8q.jpg" src="https://images.ptausercontent.com/7d0443cf-5c19-4494-98a6-0f0f54894eaa.jpg"></th> 
   <th>&nbsp;</th> 
   <th><img alt="9q.jpg" src="https://images.ptausercontent.com/d187e37a-4eb8-4215-8e2c-040a73c5c8d8.jpg"></th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>Figure 1</td> 
   <td>&nbsp;</td> 
   <td>Figure 2</td> 
  </tr> 
 </tbody> 
</table>

### Input Specification: ###

Each input file contains several test cases. The first line gives an integer K (1<K≤200). Then K lines follow, each gives a configuration in the format "N Q1 Q2... QN", where 4≤N≤1000 and it is guaranteed that 1≤Qi≤N for all i=1,⋯,N. The numbers are separated by spaces.

### Output Specification: ###

For each configuration, if it is a solution to the N queens problem, print `YES` in a line; or `NO` if not.

### Sample Input: ###

4
    8 4 6 8 2 7 1 3 5
    9 4 6 7 2 8 1 9 5 3
    6 1 5 2 6 4 3
    5 1 3 5 2 4

### Sample Output: ###

YES
    NO
    NO
    YES

我的解题思路:

由于给出了每个皇后的位置,所以我们可以从第一列开始对每个皇后进行递推,如果它的右侧行,上斜线,下斜线出现了另一个皇后,那么我们判定这不是N皇后

读入皇后坐标并在矩阵上标记为1,然后递推每一个皇后

#include <iostream>
    #include <vector>
    using namespace std;
    const int maxn = 201;
    int chessBorad[maxn][maxn];
    vector<bool> res;
    bool flag = true;
    
    int main() {
    	memset(chessBorad, 0, sizeof(chessBorad));
    	int n;
    	cin >> n;
    	for (int i = 0; i < n; ++i) {
    		int N;
    		cin >> N;
    		int a[200];
    		for (int j = 1; j <= N; ++j) {
    			int num = 0;
    			cin >> num;
    			a[j] = num;
    			chessBorad[N - num + 1][j] = 1;
    		}
    		for (int i = 1; i < N; ++i) {   //每个皇后
    			for (int j = 1; j <= N-i; ++j) {   //每个皇后距离右边界的距离
    					if (chessBorad[N - a[i] + 1][i + j] == 1)     //向右递推
    					{
    						flag = false;
    						break;
    					}
    					if (N - a[i] + 1 + j >= 1 && N - a[i] + 1 + j <= N && chessBorad[N - a[i] + 1 + j][i + j] == 1)  //斜向上递推
    					{
    						flag = false;
    						break;
    					}
    					if (N - a[i] + 1 - j >= 1 && N - a[i] + 1 - j <= N && chessBorad[N - a[i] + 1 - j][i + j] == 1)  //斜向下递推
    					{
    						flag = false;
    						break;
    					}
    			}
    			if (flag == false) {
    				break;
    			}
    		}
    		res.push_back(flag);
    		flag = true;
    		memset(chessBorad, 0, sizeof(chessBorad));
    	}
    
    	for (auto x : res) {
    		if (x) {
    			cout << "YES" << endl;
    		}
    		else {
    			cout << "NO" << endl;
    		}
    	}
    	system("PAUSE");
    	return 0;
    }