PAT 1128 N Queens Puzzle

1128 N Queens Puzzle (20 分)

 

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 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). Then K lines follow, each gives a configuration in the format "N Q​1​​ Q​2​​ ... Q​N​​", 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; 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

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    int t;
    cin >> t;
    while(t--){
        int n;
        cin >> n;
        int a[n+];
        for(int i=;i <= n;i++){
            cin >> a[i];
        }

        bool flag = ;

        for(int i=;i <= n;i++){
            // 行a[i] 列i
            for(int j=;j <= n;j++){
                if(i != j)
                    if(abs(a[i]-a[j])==abs(i-j)||a[i] == a[j]){
                        flag = ;
                        break;
                    }
            }
        }

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

    }

    return ;
}

一开始漏了行相等的情况，还有1000^2的复杂度一开始超时了？玄学测评机

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int main() {
    int k, n;
    cin >> k;
    for (int i = ; i < k; i++) {
        cin >> n;
        vector<int> v(n);
        bool result = true;
        for (int j = ; j < n; j++) {
            cin >> v[j];
            for (int t = ; t < j; t++) {
                if (v[j] == v[t] || abs(v[j]-v[t]) == abs(j-t)) {
                    result = false;
                    break;
                }
            }
        }
        cout << (result == true ? "YES\n" : "NO\n");
    }
    return ;
}

——一边输入一边就在计算了，比我这个简单点