PAT甲级——A1128 N Queens Puzzle【20】

The "eight queens puzzle" is the problem of placing eight chess queens on an 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 N non-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 (, where Q​i​​ 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.

Figure 1		Figure 2

Input Specification:

Each input file contains several test cases. The first line gives an integer K (1). Then K lines follow, each gives a configuration in the format "N Q​1​​ Q​2​​ ... Q​N​​", where 4 and it is guaranteed that 1 for all ,. 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

 #include <iostream>
 #include <vector>
 using namespace std;
 int queen[];
 int main()
 {
     int k, n, a;
     cin >> k;
     while (k--)
     {
         fill(queen, queen + , );
         cin >> n;
         bool res = true;
         for (int i = ; i <= n; ++i)
         {
             cin >> queen[i];//新一列存入queen
             for (int t = ; t < i; ++t)//判断前i-1列的queen是不是在同一行
             {
                 if (queen[i] == queen[t] || abs(queen[i] - queen[t]) == abs(i - t))//是否存在相同行，和第t列的斜线位置
                 {
                     res = false;
                     break;
                 }
             }
         }
         cout << (res == true ? "YES" : "NO") << endl;
     }
     return ;
 }