【笨方法学PAT】1128 N Queens Puzzle （20 分）_8 queens puzzle代码-优快云博客

本文链接：https://blog.youkuaiyun.com/linghugoolge/article/details/83472150

本文介绍了一个算法，用于判断给定的棋盘配置是否为有效的N皇后问题解决方案。通过检查行和对角线上的皇后冲突，该算法能够准确地确定配置是否有效。

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一、题目

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.


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 ... 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

二、题目大意

给定一串数组，判断是否满足N皇后问题。

三、考点

数组

四、注意

1、要对row进行判断；

2、要对diagonal进行判断；

3、注意，输出从0开始和皇后从1开始，要区别，否则会有一个报错。

五、代码

#include<iostream>
#include<vector>
#include<math.h>
#define N 1001
using namespace std;
int main() {
	//read
	int n;
	cin >> n;
	while (n--) {
		int m;
		cin >> m;
		vector<int> vec(m+1);
		//used for judge row
		int a[N] = { 0 };
		bool flag = true;
		for (int i = 1; i <= m; ++i) {
			int b;
			cin >> b;
			a[b]++;
			vec[i] = b;
		}

		//judge row
		for (int i = 1; i <= m; ++i) {
			if (a[i] > 1 || a[i] == 0) {
				flag = false;
				break;
			}
		}

		//judge diagonal
		for (int i = 1; i <= m; ++i) {
			for (int j = 1; j < i; ++j) {
				if (abs(i - j) == abs(vec[i] - vec[j])) {
					flag = false;
					break;
				}
			}
		}

		//output
		if (flag == false) {
			cout << "NO" << endl;
		}
		else
			cout << "YES" << endl;
	}

	system("pause");
	return 0;
}