1128 N Queens Puzzle (20 point(s))

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最新推荐文章于 2020-06-28 16:08:41 发布

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1128 N Queens Puzzle (20 point(s))

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

皇后问题局面判断。做过八皇后问题的DFS这道题就迎刃而解了。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[1005];
bool judge(int n,int a[]){
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			if(i!=j){
				if(a[i]==a[j]) return false;
				if(abs(i-j)==abs(a[i]-a[j])) return false;
			}
		}
	}
	return true;
}
int main(void){
	int N,k;cin>>N;
	while(N--){
		cin>>k;
		for(int i=0;i<k;i++) cin>>a[i];
		if(judge(k,a)) puts("YES");
		else puts("NO");
	}
	return 0;
}