PAT甲级1128

1128. N Queens Puzzle (20)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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 (Q₁, Q₂, ..., Q_N), 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 < K <= 200). Then K lines follow, each gives a configuration in the format "N Q₁ Q₂ ... Q_N", where 4 <= N <= 1000 and it is guaranteed that 1 <= Q_i <= 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

#include
#include
using namespace std;
const int maxn = 1000 + 10;
int K, N;
int a[maxn];
int main()
{
	scanf("%d", &K);
	for (int i = 0; i < K; i++)
	{
		scanf("%d", &N);
		for (int j = 0; j < N; j++)
		{
			scanf("%d", &a[j]);
		}
		bool flag = true;
		for (int j = 0; j < N; j++)
		{
			for (int k = 0; k < N; k++)
			{
				if (k != j)
				{
					if (a[k] == a[j] || abs(a[k] - a[j]) == abs(k - j))
					{
						printf("NO\n"); flag = false; break;
					}
				}
			}
			if (!flag)
				break;
		}
		if (flag)
		{
			printf("YES\n");
		}
	}
	return 0;
}