PAT A1128 N Queens Puzzle （20 分）——数学题

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 (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

 

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
int k,n;
int a[1010];
int main(){
  scanf("%d",&k);
  for(int i=0;i<k;i++){
    fill(a,a+1010,0);
    scanf("%d",&n);
    int flag=0;
    for(int j=1;j<=n;j++){
      int tmp;
      scanf("%d",&tmp);
      if(flag==0){
        a[j]=tmp;
      for(int q=1;q<j;q++){
        if(abs(j-q)==abs(a[j]-a[q]) || a[j]==a[q]){
          flag=1;
          break;
        }
      }}
      else continue;
    }
    printf("%s
",flag==1?"NO":"YES");
  }
}

注意点：测试点1是有两个在同一行，所以这题是不仅判断对角线，还有同行，但肯定不会同列。