1128 N Queens Puzzle （20 分）

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本文介绍了一个算法，用于判断给定的棋盘配置是否为N皇后问题的有效解决方案。通过检查每行、每列以及对角线上的皇后是否冲突，来确定配置是否合法。

1128 N Queens Puzzle （20 分）

The "eight queens puzzle" is the problem of placing eight chess queens on an 8×88\times 88×8 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 NNN queens problem of placing NNN non-attacking queens on an N×NN\times NN×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,⋯,QN)(Q_1, Q_2, \cdots , Q_N)(Q1,Q2,⋯,QN), where QiQ_iQi is the row number of the queen in the iii-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 KKK (1<K≤2001<K\le 2001<K≤200). Then KKK lines follow, each gives a configuration in the format "NNN Q1Q_1Q1 Q2Q_2Q2 ... QNQ_NQN", where 4≤N≤10004\le N\le 10004≤N≤1000 and it is guaranteed that 1≤Qi≤N1\le Q_i\le N1≤Qi≤N for all i=1,⋯,Ni=1, \cdots , Ni=1,⋯,N. The numbers are separated by spaces.

Output Specification:

For each configuration, if it is a solution to the NNN 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<iostream>
#include<vector>
#include<cmath>
using namespace std;
int main(){
	int k,n;
	cin >> k;
	while(k--){
		int flag = 0;
		cin >> n;
		int a[1001]={0};
		vector<int> v(n+1,0);
		for(int i = 0;i < n;i++){
			cin >> a[i];
			if(v[a[i]] == 1){
				flag = 1;
			}
			v[a[i]] = 1;
		}
		for(int i = 0;i < n-1;i++){
			for(int j = i+1;j < n;j++){
				if(abs(a[i]-a[j])==j-i){
					flag = 1;	
					break;
				}
			}
		}
		if(flag){
			cout << "NO" << endl;
		}else{
			cout << "YES" << endl;
		}
	}
	return 0;
}