题目描述：
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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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: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 4Sample Output:
YES NO NO YES
已知条件：
1.经典的8皇后问题是让求出8个皇后的位置，怎样能使得她们不想不在同一行，同一列，以及同一对角线。
2.题目是给出N皇后的一种解，判断该解是否合理。
3.有K组测试用例
4.在每个K行中，第一个数字N表示N个皇后，接下来Q1,Q2,Q3...Qi...QN表示皇后在Qi行i列。
解决方法：
1.这里我们声明一个数组c[3][2*MAXN]
2.c[0][i],c[1][i],c[2][i]分别表示i行和左右对角线是否被占据
3.对角线可以用行坐标±列坐标得到
注意：
因为减法可能涉及到负数，所以在做减法的时候做“+n”处理。
#include<cstdio>
#include<cstring>
#define MAXN 1001
using namespace std;
int a[MAXN], n, t;
int c[3][2*MAXN];
int main()
{scanf("%d",&t);while(t--){memset(c,0,sizeof(c));scanf("%d",&n);int flag = 1;for(int i=1 ;i<=n ;i++){scanf("%d",&a[i]);if(c[0][a[i]]||c[1][a[i]-i+n]||c[2][a[i]+i]) flag = 0;c[0][a[i]] = 1; //row c[1][a[i]-i+n] = 1; //diagonal c[2][a[i]+i] = 1; //diagonal }if(flag)puts("YES");else puts("NO");}return 0;
}