Source:

Description:

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 Klines 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:

48 4 6 8 2 7 1 3 59 4 6 7 2 8 1 9 5 36 1 5 2 6 4 35 1 3 5 2 4

Sample Output:

YESNONOYES

Keys:

• 简单模拟

Attention:

• 给出的N个皇后可能在同一行

Code:

1 /* 2 Data: 2019-05-25 10:24:05 3 Problem: PAT_A1128#N Queens Puzzle 4 AC: 18:04 5  6 题目大意： 7 判断给定的序列是否为N皇后问题的解 8 输入： 9 第一行给出：测试数K<=200;10 接下来K行，问题规模N<=1e3，N个数11 输出：12 Yes or No13 */14 15 #include
     
      16 #include
      
       17 using namespace std;18 const int M=1e3+10;19 20 int main()21 {22 #ifdef    ONLINE_JUDGE23 #else24     freopen("Test.txt", "r", stdin);25 #endif26 27     int n,m;28     scanf("%d", &m);29     while(m--)30     {31         scanf("%d", &n);32         int q[M],ans=1;33         for(int i=1; i<=n; i++)34             scanf("%d", &q[i]);35         for(int i=1; i<=n; i++)36         {37             for(int j=i+1; j<=n; j++)38                 if(abs(j-i)==abs(q[j]-q[i]) || q[j]==q[i]){39                     ans=0;break;40                 }41             if(ans==0)42                 break;43         }44         if(ans) printf("YES\n");45         else    printf("NO\n");46     }47 48     return 0;49 }