PAT 甲级 1128 N Queens Puzzle

https://pintia.cn/problem-sets/994805342720868352/problems/994805348915855360

The "eight queens puzzle" is the problem of placing eight chess queens on an 8×8chessboard 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 (Q​1​​,Q​2​​,⋯,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 Q​1​​ Q​2​​ ... 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.

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

时间复杂度：$O(1/2 * n^2)$ 

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
int a[1111], line[1111], row[1111];

int main() {
    int K;
    scanf("%d", &K);
    while(K --) {
        scanf("%d", &N);

        for(int i = 1; i <= N; i ++) {
            scanf("%d", &a[i]);
        }

        memset(line, 0, sizeof(line));
        memset(row, 0, sizeof(row));

        for(int i = 1; i <= N; i ++) {
            line[a[i]] ++;
            row[i] ++;
        }

        bool flag = true;
        for(int i = 1; i <= N; i ++) {
            if(line[i] != 1 || row[i] != 1)
            flag = false;
        }

        for(int i = 2; i <= N; i ++) {
            for(int j = 1; j < i; j ++) {
                if(abs(a[i] - a[j]) == abs(i - j))
                    flag = false;
            }
        }

        if(flag)
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}