PAT甲级 1128. N Queens Puzzle (20)

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

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题目的意思是给出n列皇后位置，判断是否不会互相攻击

思路：n^2暴力判断

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[10005];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        int flag=0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<i; j++)
            {
                if(a[i]==a[j]||fabs(a[i]-a[j])==fabs(j-i))
                {
                    flag=1;
                    break;
                }
            }
        printf("%s\n",flag==1?"NO":"YES");
    }
    return 0;
}