1128. N Queens Puzzle (20)

1128. N Queens Puzzle (20)

時間限制

300 ms

內存限制

65536 kB

代碼長度限制

16000 B

判題程序

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

題意：判斷八皇后——給出皇后的位置，判斷皇后相互之間是否交叉。PS:皇后可以 上下左右 和 斜上下左右 任意走。

思路：判斷橫座標相差N個單位的兩個皇后，縱座標是否也相差N個單位，是則爲NO。 

           另外還要判斷皇后是否在一條直線上   _(:з」∠)_開始做的時候忘了這茬，bug半天看不出來

代碼：

#include<iostream>
#include<string>
using namespace std;
int check(int n)
{
	int num[1010];
	for (int i = 0; i < n; i++)
		cin >> num[i];
	for (int i = 0; i < n; i++)
		for (int j = i + 1; j < n; j++)
			if (abs(num[i] - num[j]) == abs(i - j) || num[i] == num[j]) return 0;
	return 1;
}
int main()
{
	int k, n;
	cin >> k;
	while (k--)
	{
		cin >> n;
		if (check(n)) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
}