PAT 1128 N Queens Puzzle

最新推荐文章于 2025-07-29 21:31:10 发布

韵途琼梦

最新推荐文章于 2025-07-29 21:31:10 发布

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文章标签： c++ 算法数据结构 pat考试

本文链接：https://blog.youkuaiyun.com/weixin_60830668/article/details/133023211

本文介绍了一个编程题目，要求判断给定的棋盘配置是否满足N皇后问题的条件，即不使任何两个皇后互相威胁。程序通过检查行、列和对角线来确定答案。

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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.


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:

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

Code:

#include<bits/stdc++.h>
using namespace std;
const int N=1100;
int n;
bool row[N],udg[N*2],dg[N*2];
int main(){
    cin>>n;
    while(n--){
        memset(row,0,sizeof row);
        memset(udg,0,sizeof udg);
        memset(dg,0,sizeof dg);
        int k;
        cin>>k;
//        cout<<k<<endl;
        bool flag=true;
        for(int i=1;i<=k;i++){
            int x;
            cin>>x;
           if(!udg[i-x+k]&&!dg[x+i]&&!row[x]){
               udg[i-x+k]=true;
               dg[x+i]=true;
               row[x]=true;
           }else{
               flag=false;

           }
        }
        if(flag)
            puts("YES");
        else
            puts("NO");
    }
}