1128. N Queens Puzzle (20)

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本文介绍了一个简化版的N皇后问题，任务是判断给定的棋盘配置是否为有效解。通过建立坐标系并检查三个方向的映射来验证棋盘状态的有效性。

1128. N Queens Puzzle (20)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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, ..., Q_N), 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 < K <= 200). Then K lines follow, each gives a configuration in the format "N Q₁ Q₂ ... Q_N", where 4 <= N <= 1000 and it is guaranteed that 1 <= Q_i <= 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：建立坐标系，检查三次映射，分别是点在y轴上的映射（数值是y）、向左45度在x轴上的映射（数值是x-y）、向右45度在x轴上的映射（数值是x+y）

只要有一次重合则表示棋盘违规，下图图二画圈部分重合（即x-y均为-5）

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>

using namespace std;

int main()
{
    //freopen("in.txt","r",stdin);

    int K;
    cin>>K;
    while(K--)
    {
        int book1[2000]={0};
        int book2[2000]={0};
        int book3[2000]={0};
        int N,i,x,y,flag=1;
        cin>>N;
        for(i=1;i<=N;i++)
        {
            int t;
            cin>>t;
            x=i;
            y=t;

            if(book1[y]==0)
                book1[y]++;
            else
                flag=0;

            if(book2[x-y]==0)
                book2[x-y]++;
            else
                flag=0;

            if(book3[x-y]==0)
                book3[x-y]++;
            else
                flag=0;
        }

        if(flag==0)
            printf("NO\n");
        else
            printf("YES\n");

    }


    return 0;
}