PAT_A1128#N Queens Puzzle

Source:

PAT A1128 N Queens Puzzle (20 分)

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:

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

Keys:

• 简单模拟
• 八皇后问题

Attention:

• 给出的N个皇后可能在同一行

Code:

 /*
 Data: 2019-05-25 10:24:05
 Problem: PAT_A1128#N Queens Puzzle
 AC: 18:04

 题目大意：
 判断给定的序列是否为N皇后问题的解
 输入：
 第一行给出：测试数K<=200;
 接下来K行，问题规模N<=1e3，N个数
 输出：
 Yes or No
 */

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int M=1e3+;

 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif

     int n,m;
     scanf("%d", &m);
     while(m--)
     {
         scanf("%d", &n);
         int q[M],ans=;
         for(int i=; i<=n; i++)
             scanf("%d", &q[i]);
         for(int i=; i<=n; i++)
         {
             for(int j=i+; j<=n; j++)
                 if(abs(j-i)==abs(q[j]-q[i]) || q[j]==q[i]){
                     ans=;break;
                 }
             if(ans==)
                 break;
         }
         if(ans) printf("YES\n");
         else    printf("NO\n");
     }

     return ;
 }