2017浙江大学计算机和软件考研复试题A题 N Queens Puzzle
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, ..., QN), where Qi 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 |
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 ... QN", where 4 <= N <= 1000 and it is guaranteed that 1 <= Qi <= 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:
Sample Output:
解析：这道题的大概意思是给你一个皇后问题的一种情况，让你判断是不是皇后问题的一个解，如果是输出YES，否则输出NO。
这道题作为复试的第一题肯定要送点分，所以规定了每一列只会有一个皇后，那么只要判断每个皇后的同一行和对角线上有没有另一个皇后就行了。
判断同一行是否有两个皇后很简单，至于对角线，只要判断两个皇后所在行的差的绝对值是否等于列之差就能解决。
代码如下：
#include<iostream>
using namespace std;
int k;
bool check(int*input,int N) {
for (int i = 1; i <= N; i++) {
for (int j = i + 1; j <= N; j++) {
if (input[j] == input[i]) return false;//两个皇后在同一行
else if (abs(input[j] - input[i]) == (j - i)) return false;//两个皇后在对角线
}
}
return true;
}
int main() {
cin >> k;
for (int Case = 0; Case < k; Case++) {
int N;
cin >> N;
int*input;
input = new int[N + 1];
for (int i = 1; i <= N ; i++) cin >> input[i];
if (check(input, N)) cout << "YES" << endl;
else cout << "NO" << endl;
}
}