PAT甲級 1128

題目 N Queens Puzzle

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 (Q​1​​ ,Q​2​​ ,⋯,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.
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 Q​1​​ Q​2​​ … 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.

解析

判斷給出的數據是否滿足皇后問題

代碼

#include<bits/stdc++.h>
#include<string>

#define INF 1<<29

using namespace std;

struct node {
    int x, y;
};

void pat1128() {
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        vector<node> data;
        int t;
        cin >> t;
        for (int j = 1; j <= t; ++j) {
            int num;
            cin >> num;
            data.push_back({j, num});
        }
        bool flag = true;
        for (int j = 0; j < data.size() ; ++j) {
            for (int k = j + 1; k < data.size(); ++k) {
                if (data[j].x == data[k].x || data[j].y == data[k].y ||
                    abs(data[j].x - data[k].x) == abs(data[j].y - data[k].y)) {
                    flag = false;
                    break;
                }
            }
        }
        if (flag) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }

}

int main() {
    pat1128();
    return 0;
}