1128. N Queens Puzzle (20)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (Q₁, Q₂, ..., 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 Q₁ Q₂ ... 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:

48 4 6 8 2 7 1 3 59 4 6 7 2 8 1 9 5 36 1 5 2 6 4 35 1 3 5 2 4

Sample Output:

YESNONOYES 思路 N皇后问题，只不过题目仅要求判断是否是同一行和同一对角线。 对于每一个输入的数nums[i]。 1.判断是否和前面的皇后棋子是否在同一行,即输入的第i个数和前i-1个数是否相同。 2.判断是否在同意对角线上，即第i个数和前i-1个数的斜率是否相同，即abs(nums[i] - nums[k]) == abs(i - k) ( 0 <= k < i)是否成立。 代码

#include
      
       #include
       
        #include
        
         using namespace std;int main(){   int K;   while(cin >> K)   {       for(int i = 0;i < K;i++)       {           int N;           cin >> N;           vector
         
           nums(N);           bool isSolution = true;           for(int j = 0;j < N;j++)           {               cin >> nums[j];               for(int k = 0;k < j;k++)               {                 if(nums[k] == nums[j] || abs(nums[j] - nums[k]) == abs(j - k))                 {                     isSolution = false;                     break;                 }               }           }           if(isSolution)                cout << "YES" << endl;               else                cout << "NO" << endl;       }   }}