1128 N Queens Puzzle (20 分)

The "eight queens puzzle" is the problem of placing eight chess queens on an 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×Nchessboard. (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 (, 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). Then K lines follow, each gives a configuration in the format "N Q​1​​ Q​2​​ ... Q​N​​", where 4 and it is guaranteed that 1 for all ,. 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

判断n皇后是否正确
只要判断列，然后对角线就可以。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n,m,x;
 4 int an[2005],bn[2005],cn[2005];
 5 int main(){
 6     cin >> n;
 7     while(n--){
 8         cin >> m;
 9         memset(an,0,sizeof(an));
10         memset(bn,0,sizeof(bn));
11         memset(cn,0,sizeof(cn));
12         bool flag = true;
13         for(int i = 1; i <= m; i++){
14             cin >> x;
15             int a = i-x+1000, b = x-i+1000;
16             if(an[x] == 1 || bn[a] == 1 || cn[b] == 1)
17                 flag = false;
18             else
19                 an[x] = 1, bn[a] = 1, cn[b] = 1;
20         }
21         if(flag){
22             cout <<"YES"<<endl;
23         }else{
24             cout <<"NO"<<endl;
25         }
26     }
27     return 0;
28 }