PAT 1128 N Queens Puzzle (20)

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.

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  Figure 1 Figure 2

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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

额，没想到暴力膜居然通过了，三层循环嵌套，啧啧；
思路：对于合法的放置， 任意两个棋子不在同一行， 或者连线的斜率不为1或者-1； 如果存在则输出NO， 反之输出YES

#include<iostream>
#include<vector>
using namespace std;
int main(){
  int n, k, i, j;
  cin>>n;
  for(i=1; i<=n; i++){
    cin>>k;
    vector<int> v(k+1);
    bool flag=true;
    for(j=1; j<=k; j++){
      scanf("%d", &v[j]);
      for(int y=1; y<j && flag; y++){
        if(v[j]-v[y]==j-y || v[j]-v[y]==y-j || v[j]==v[y]){
          cout<<"NO"<<endl;
          flag=false;
        }
      }
    }
    if(flag) cout<<"YES"<<endl;
  }
  return 0;
}