1128 N Queens Puzzle-PAT甲级

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本文介绍了一个简化版的八皇后问题，即如何判断一个给定的棋盘配置是否为有效的解决方案。通过检查每一对皇后是否在同一行或对角线上，算法能够快速判断配置的有效性。

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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

判断任何两个是否在同一行或对角线上

如果两个点的横坐标差的绝对值等于纵坐标差的绝对值，则二者在同一对角线上

满分代码如下

#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int graph[N],vst[N];
int k,n;
bool check(){
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			if(graph[i]==graph[j]||(abs(j-i)==abs(graph[j]-graph[i])))
				return false;
		}
	}
	return true;
}
int main(){
	cin>>k;
	for(int i=1;i<=k;i++){
		cin>>n;
		for(int j=1;j<=n;j++){
			cin>>graph[j];
		}
		if(check()){
			cout<<"YES"<<endl;
		}else{
			cout<<"NO"<<endl;
		}
	}
	return 0;
}