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1128 N Queens Puzzle (20分)
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.
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
本题思路:对于每一组数据,用几个数组来记录数据不重复即可,因为是按列排序的,因此只需要保证行(row)不重复,主对角线(col+row)不重复,反对角线(col-row)不重复即可,(需要注意的是计算反对角线时可能会出现col-row<0的情况下,而数组不存在下标为负数的情况,因此只需整体右移N个单位即可)
AC代码:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;int main()
{int n;cin >> n;while(n--){bool flag=true;int k;cin >> k;vector<int> row(k+1,0);vector<int> dia(2*k+2,0);vector<int> backdia(2*k+2,0);vector<int> list(k+1,0);for(int i=1;i<=k;i++){cin >> list[i];row[list[i]]++;if(row[list[i]]>1)flag=false;dia[i+list[i]]++;if(dia[i+list[i]]>1)flag=false;backdia[i-list[i]+k]++;if(backdia[i-list[i]+k]>1)flag=false;}if(flag)printf("YES\n");elseprintf("NO\n");}return 0;
}
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