PAT-1128 N Queens Puzzle

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本文介绍了一个程序，用于判断给定的N皇后配置是否为有效解。通过读取多个测试用例，程序检查每种配置是否满足N皇后问题的要求，即没有两个皇后在同一行、同一列或同一对角线上。

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1128 N Queens Puzzle（20 分）

The "eight queens puzzle" is the problem of placing eight chess queens on an 8×8chessboard 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×Nchessboard. (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 Qiis 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

判断N皇后的摆法是否合法，主要是两条对角线的判断。

#include<stdio.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
using namespace std;
#define NUM 2005

int main(){
	int k,n;
	scanf("%d",&k);
	for(int i=0;i<k;i++){
		int n;
		scanf("%d",&n);

		int t;
		bool lt[NUM]={0};
		bool rt[NUM]={0};
		bool row[NUM]={0};

		for(int j=0;j<n;j++){
			scanf("%d",&t);
			row[t]=true;
			lt[j+t]=true;
			rt[j+1+n-t]=true;
		}
		bool ok=true;
		for(int j=1;j<=n;j++){
			if(!row[j]){ok=false;break;}
		}
		int count=0;
		for(int j=1;j<=n*2-1;j++){
			if(lt[j])count++;
		}
		if(count!=n)ok=false;
		count=0;
		for(int j=1;j<=n*2-1;j++){
			if(rt[j])count++;
		}
		if(count!=n)ok=false;

		if(ok)printf("YES\n");
		else printf("NO\n");
	}

	return 0;
}