1128 N Queens Puzzle

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#c++ #算法 #pat考试

本文介绍了一个简化版的八皇后问题，即不需求解而是判断给定棋盘配置是否为有效解。通过输入一系列棋盘配置，利用C++程序判断其是否满足八皇后问题的要求，即任意两个皇后不在同一行、列或对角线上。

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, …, QN), where Qi 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.

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 … QN”, where 4 <= N <= 1000 and it is guaranteed that 1 <= Qi <= 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

#include<iostream>
#include<vector>
#include<string.h>
using namespace std;
int main(){
	int c,n,i,j,cur;
	int arr1[2005],arr2[2005],arrc[1005],arrr[1005];
	cin>>c;
	for(i = 0;i<c;i++){
		memset(arr1,0,8020);
		memset(arr2,0,8020);
		memset(arrc,0,4020);
		memset(arrr,0,4020);
		cin>>n;
		bool ret = true;
		for(j = 1;j <= n;j++){
			cin>>cur;
			if(ret){
				if(arr1[j + cur] == 0)arr1[j + cur]++;
				else ret = false;
				if(arr2[j - cur + 1000] == 0)arr2[j - cur + 1000]++;
				else ret = false;
				if(arrc[j] == 0)arrc[j]++;
				else ret = false;
				if(arrr[cur] == 0)arrr[cur]++;
				else ret = false;
			}
		}
		if(ret)cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}