PAT (Advanced Level) Practise 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:

48 4 6 8 2 7 1 3 59 4 6 7 2 8 1 9 5 36 1 5 2 6 4 35 1 3 5 2 4

Sample Output:

YESNONOYES

题意：k组数据，每组数据先给你n个皇后，然后每列皇后行的位置，问这样放的n个皇后是否满足任意两个皇后不在同一行，同一列，同一斜线上

解题思路：记录下每行出现的次数，然后判断是不是在同一斜线上

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, x[1005], visit[1005];int main(){int t;scanf("%d", &t);while (t--){scanf("%d", &n);memset(visit, 0, sizeof visit);int flag = 1;for (int i = 1; i <= n; i++){scanf("%d", &x[i]);if (visit[x[i]]) flag = 0;visit[x[i]] = 1;}if (!flag) { printf("NO\n"); continue; }for (int i = 2; i <= n; i++)for (int j = 1; j < i; j++)if (abs(i - j) == abs(x[i] -x[j])) { flag = 0; break; }if (flag) printf("YES\n");else printf("NO\n");}return 0;}