PAT 甲级 1128 N Queens Puzzle
https://pintia.cn/problem-sets/994805342720868352/problems/994805348915855360
The "eight queens puzzle" is the problem of placing eight chess queens on an 8×8chessboard 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.
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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 ... 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
#include <bits/stdc++.h> using namespace std; int N; int a[1111], line[1111], row[1111]; int main() { int K; scanf("%d", &K); while(K --) { scanf("%d", &N); for(int i = 1; i <= N; i ++) { scanf("%d", &a[i]); } memset(line, 0, sizeof(line)); memset(row, 0, sizeof(row)); for(int i = 1; i <= N; i ++) { line[a[i]] ++; row[i] ++; } bool flag = true; for(int i = 1; i <= N; i ++) { if(line[i] != 1 || row[i] != 1) flag = false; } for(int i = 2; i <= N; i ++) { for(int j = 1; j < i; j ++) { if(abs(a[i] - a[j]) == abs(i - j)) flag = false; } } if(flag) printf("YES\n"); else printf("NO\n"); } return 0; }