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 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.

8q.jpg   9q.jpg
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 "Q1​​ Q2​​ ... QN​​", 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:

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 No
13 */
14 
15 #include<cstdio>
16 #include<algorithm>
17 using namespace std;
18 const int M=1e3+10;
19 
20 int main()
21 {
22 #ifdef    ONLINE_JUDGE
23 #else
24     freopen("Test.txt", "r", stdin);
25 #endif
26 
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 }

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转载自www.cnblogs.com/blue-lin/p/10921801.html