PAT 1128 N Queens Puzzle（20 分）

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

思路：判断有没有存在任何两个皇后在同一对角线、反对角线或同行同列；

程序：

#include <cstdio>
#include <iostream>
#include <vector>
using namespace std;
bool isvalid(vector<int> v)
{
  for(int i = 0; i < v.size(); i++)
  {
    for(int k = 1; k + i < v.size(); k++)
    {
      if(v[i] == v[k+i] - k || v[i] == v[i+k] + k || v[i] == v[k+i])
        return false;
    }
  }
  return true;
}
int main()
{
  int n;
  scanf("%d",&n);
  while(n--)
  {
    vector<int> v;
    int m;
    scanf("%d",&m);
    for(int i = 0; i < m; i++)
    {
      int a;
      scanf("%d",&a);
      v.push_back(a);
    }
    if(isvalid(v))
      printf("YES\n");
    else
      printf("NO\n");
  }
}