PAT--1128. 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 (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.

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

题解

对于一个n∗n的棋盘，可以用 chess[3][n * 2] 分别标记一行、一列以及对角线是否出现相同棋子。

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;const int maxn = 1000 + 5;int n, k, pos;int chess[3][2 * maxn];int main(){#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);#endif // ONLINE_JUDGE    cin >> k;    while(k--){        cin >> n;        int flag = 1;        memset(chess, 0, sizeof(chess));        for(int i = 1; i <= n; ++i){            cin >> pos;            if(chess[0][pos] || chess[1][pos + i] || chess[2][pos - i + n]) flag = 0;            chess[0][pos] = 1;            chess[1][pos + i] = 1;            chess[2][pos - i + n] = 1;        }        cout << (flag ? "YES" : "NO") << endl;    }    return 0;}