PAT_A1128#N Queens Puzzle
Source:
Description:
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×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 (, 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.
Figure 1 Figure 2
Input Specification:
Each input file contains several test cases. The first line gives an integer K (1). Then Klines follow, each gives a configuration in the format "N Q1 Q2 ... QN", 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; orNO
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
Keys:
- 简单模拟
- 八皇后问题
Attention:
- 给出的N个皇后可能在同一行
Code:
/*
Data: 2019-05-25 10:24:05
Problem: PAT_A1128#N Queens Puzzle
AC: 18:04 题目大意:
判断给定的序列是否为N皇后问题的解
输入:
第一行给出:测试数K<=200;
接下来K行,问题规模N<=1e3,N个数
输出:
Yes or No
*/ #include<cstdio>
#include<algorithm>
using namespace std;
const int M=1e3+; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif int n,m;
scanf("%d", &m);
while(m--)
{
scanf("%d", &n);
int q[M],ans=;
for(int i=; i<=n; i++)
scanf("%d", &q[i]);
for(int i=; i<=n; i++)
{
for(int j=i+; j<=n; j++)
if(abs(j-i)==abs(q[j]-q[i]) || q[j]==q[i]){
ans=;break;
}
if(ans==)
break;
}
if(ans) printf("YES\n");
else printf("NO\n");
} return ;
}
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