1128 N Queens Puzzle

1128 N Queens Puzzle
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 (Q
1
,Q
2
,⋯,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.

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<K≤200). Then K lines follow, each gives a configuration in the format "N Q
1
Q
2
… 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
题目非常有意思！
分析：n皇后问题，所有的棋子都要处在不同的行与不同的对角线上。
判断是否位于同一行非常重要。
如何判断是否位于同一对角线呢？
观察可发现，假设棋子的坐标（x，y),x-y处在确定的集合中。例如八皇后问题，x-y 是集合{7，6，5，4，3，2，1，0，-1，-2，-3，-4，-5，-6，-7}中的元素。现在有思路了。
参考代码：

#include<vector>
#include<iostream>
#include<cmath>
using namespace std;

int main() {
	int k, n, flag,flag2,temp;
	scanf_s("%d", &k);
	for (int i = 1; i <= k; i++) {
		scanf_s("%d", &n);
		flag = 1; flag2 = 1;
		vector<int>a(n + 1, 0);
		vector<int>b(2 * n+2, 0);
		for (int j = 1; j <= n; j++) {
			scanf_s("%d", &temp);
			if (a[temp] == 0)a[temp] = 1;
			else flag = 0;
			if (b[temp - j + n] == 0)b[temp - j + n] = 1;
			else flag2 = 0;
		}
		if (flag&&flag2)cout << "YES" << endl;
		else cout << "NO" << endl;
	}

	return 0;
}

猜你喜欢