「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  , where  ​ 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  . Then K lines follow, each gives a configuration in the format “ ​”, where 4≤N≤1000 and it is guaranteed that  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

Ω

给出几个棋盘布局，判断是否满足“八皇后问题”的要求，即行列以及对角线上最多只能有一个皇后。棋盘布局的表示方法是给出每一列皇后所在的(自下而上)行数。

题目保证了棋盘上的每一列均有一个皇后，因此我们只要检查行与对角线之间是否满足。行检查比较显然，只要所有皇后的行数不同；而对于对角线的条件，则需要满足 ，其中 是列数， 表示 列皇后所在的行数。每一列的皇后只需和前面几列的皇后比较即可。

🐎

#include <iostream>
#include <vector>using namespace std;int main()
{int n, m;cin >> n;for (int i = 0; i < n; ++i){cin >> m;vector<int> seq(m);bool ans = true;for (int j = 0; j < m; ++j){cin >> seq[j];if (!ans) continue;for (int k = 0; k < j; ++k)ans = ans && (j - k != abs(seq[j] - seq[k])) && seq[j] != seq[k];}printf(ans ? "YES\n" : "NO\n");}
}