PAT 1128 N Queens Puzzle （20 分）等腰三角形的应用燚

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

题目大意：给出一个n*n的棋盘，上面放上n个棋子，如果每一个棋子都不在同一列，同一行，同一斜线上，则输出“YES”否则输出“NO” 每一次有多组输入。

思路：1.由于输入的数列array中任意array[i]=m表示该棋子放在第m行的第i列，根据数组下标递增的性质可知这n棋子在不同的列上，所以只需判断棋子是否在同一行，或者在同一斜线上。

2.为了降低复杂度，我们采用边输入边判断的方法，即判断已输入的棋子是否满足题意。我们用一个bool值flag=false来标记输入的数列是否满足题意，如果array[i]==array[j]则这两颗棋子在同一行，flag=true，如果array[i]和array[j]在同一斜线上，他们可以构成一个等腰直角三角形，所在行的差值等于所在列的差值。

难点：判断两个棋子是否在同一斜线上，解决方案利用等腰直角三角形，如果array[i]和array[j]在同一斜线上，他们可以构成一个等腰直角三角形，所在行的差值等于所在列的差值

#include<iostream>
#include<vector>
#include<math.h>
using namespace std;
int main() {
	int k;
	cin >> k;
	while (k--) {
		int n;
		cin >> n;
		bool flag = false;
		vector<int>que(n + 1, 0);//记录输入的棋子坐标，角标为列坐标，内容为横坐标
		for (int i = 1;i <= n;i++) {
			cin >> que[i];
			if (i > 1) {
				for (int j = 1;j < i;j++) {
				  //判断是否在同一行
					if (que[j] == que[i]) {
						flag = true;
						break;
					}
					//判断是否在同一斜线上
					else {
						if (abs(j - i) == abs(que[j] - que[i])) {
							flag = true;
							break;
						}	
					}
				}
			}
		}
		if (flag)
			cout << "NO" << endl;
		else
			cout << "YES" << endl;
		}
	}