PAT (Advanced Level) Practice — 1128 N Queens Puzzle （20 分）

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805348915855360

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.

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

：题意 判断不同行，不同列（已经保证了），不在对角线上。

#include<iostream>
#include<cmath>
using namespace std;
int main(){
	int T,n;
	int a[1010];
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			cin>>a[i];
		} 
		int f=0;
		for(int i=1;i<=n;i++)
		{
			if(f)break;
			for(int j=i+1;j<=n;j++)
			{
				if(f)break;
				if(a[i]==a[j])f=1;
				if(abs(i-j)==abs(a[i]-a[j]))f=1;
			}	
		} 
		if(f){
			printf("NO\n");
		}else{
			printf("YES\n");
		}
	}
}