A1128 N Queens Puzzle (20 point(s))

N皇后--判断对角线上行差值==列差值

1. 原文

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

2. 解析

N皇后，放置的皇后每行每列都不重复

判断式 abs(j-i)==abs(num[j]-num[i]) || num[j]-num[i]

3. AC代码

#include<cstdio>
#include<algorithm>
using namespace std;
int num[1010]={};
int main()
{
	int k;
	scanf("%d",&k);
	while(k--)
	{
		int n;
		scanf("%d",&n);
		int a; int flag=0;
		for (int i=0;i<n;i++)
		{
			scanf("%d",&num[i]);
			if (flag==1)
			{
				continue;
			}
			for (int j = 0; j < i; ++j)
			{
				if ((num[j]==num[i])||(abs(j-i)==abs(num[j]-num[i])))
				{
					flag=1;
					break;
				}
			}
		}
		if (flag==0)
		{
			printf("YES\n");
		}else{
			printf("NO\n");
		}
	}
	return 0;
}