1128 N Queens Puzzle （20 分）

最新推荐文章于 2020-03-15 08:47:19 发布

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最新推荐文章于 2020-03-15 08:47:19 发布

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分类专栏： PAT甲级(Advanced Level) 文章标签： PAT

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本文介绍了一种新颖的数学方法来判断一个给定的棋盘配置是否解决了八皇后问题或更一般的N皇后问题。通过分析棋盘上的皇后位置，确保它们不在同一行、列或对角线上，从而验证解决方案的有效性。

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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

思路：这里给出一种不一样的数学解法：八皇后问题中的摆放位置（x,y） x-y相同为同一条主对角线，x+y相同为同一副对角线。

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<unordered_map>
#include<queue>
#include<algorithm>
#include<cmath>
using namespace std;
#define ll long long
#define vi vector<int>
#define pii pair<int, int>
#define x first
#define y second
const int maxn=1111;
int ans[maxn],isok[maxn]={0},isokf[maxn]={0};
set<int> sett,sett1,settc;
int main()
{
  int k,n;
  scanf("%d",&k);
  for(int i=0;i<k;i++)
  {
      scanf("%d",&n);
      for(int i=1;i<=n;i++)
      {
          scanf("%d",&ans[i]);
          isok[i]=ans[i]-i;
          isokf[i]=ans[i]+i;
          settc.insert(ans[i]);
      }
      for(int i=1;i<=n;i++)
      {
          sett.insert(isok[i]);
          sett1.insert(isokf[i]);
      }
      if(sett.size()<n || sett1.size()<n||settc.size()<n)
      {
          cout<<"NO"<<endl;
      }
      else{

          cout<<"YES"<<endl;
      }
      sett.clear();
      sett1.clear();
      settc.clear();
      memset(isok,0,sizeof(isok));
      memset(isokf,0,sizeof(isokf));
      memset(ans,0,sizeof(ans));
  }
  return 0;
}