1128. N Queens Puzzle (20) 判断是否是对角线

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.

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

YES
NO
NO
YES

#include <iostream>
#include <vector>
using namespace std;

//abs()：取绝对值后取整   fabs()：直接取绝对值
int main(){
    int k, n;
    cin >> k;
    for (int i = 0; i < k; i++) {
        cin >> n;
        vector<int> v(n);//直接开int数组 测试点3会快1ms  考虑到有好几台服务器 还是忽略这一点吧
        bool result = true;
        for (int j = 0; j < n; j++) {
            cin >> v[j];
            for (int l = 0; l < j; l++) {
                if (v[j] == v[l] || abs(v[j] - v[l]) == abs(j - l)) {
                    result = false;
                }
            }
        }
        cout << (result == true ? "YES" : "NO") << endl;
    }
    return 0;
}

        

          
        
        
        

          

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