PAT 1128 N Queens Puzzle

The "eight queens puzzle" is the problem of placing eight chess queens on an 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×Nchessboard. (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 (, 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.

Input Specification:

Each input file contains several test cases. The first line gives an integer K (1). Then K lines follow, each gives a configuration in the format "N Q​1​​ Q​2​​ ... Q​N​​", where 4 and it is guaranteed that 1 for all ,. 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

　　

结果

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
//const int maxk = 210;
const int maxn = 1010;
int k,n,a[maxn] ={0};

bool isOK(int a[], int n){
    bool flag = true;
    for(int i=1;i<n;i++){
        for(int j=i+1;j<=n;j++){
            if(a[i] == a[j] or abs(i-j) == abs(a[i]-a[j])){
                flag = false;
                return flag;
            }
        }
    }
    return flag;
}

int main(){
    scanf("%d",&k);
    for(int i=0;i<k;i++){
        // input data;
        scanf("%d",&n);
        for(int j=1;j<=n;j++){
            scanf("%d",&a[j]);
        }
        // get result,and print result
        bool result;
        result = isOK(a,n);
        if(result) printf("YES");
        else printf("NO");
        if(i<k-1) printf("
");
    }
    return 0;
}

备注：
1）写的时候，我发现在PAT考试中，可以取个巧；不用安装输入完成后再全部输出的模式进行。也可以输入一行立刻输出一行结果。也不影响最终的评判。