1128 N Queens Puzzle (20分)

题目地址

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 (Q
​1
​​ ,Q
​2
​​ ,⋯,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 Q
​1
​​ Q
​2
​​ ... 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:
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

作者: CHEN, Yue
单位: 浙江大学
时间限制: 300 ms
内存限制: 64 MB

#include <cstdio>
#include <cstring>
#include<iostream>
#include <vector>
#include<math.h>
#include<string>
#include<queue>
#include <algorithm>
#include<set>
#include<map>
using namespace std;
const int  maxn= 1010;  //最大顶点数
int main(){
    int n,k,row;
    cin>>k;
    for(int count=0;count<k;count++){
        int q[maxn]={0},res=1;
        cin>>n;
        for(int col=1;col<=n;col++){
            cin>>q[col];
            for(int i=1;i<col;i++){
                if(q[col]==q[i]||abs(q[col]-q[i])==abs(col-i) )         //只要有一列的row值与col差为1，则在对角线；
                res=0;
            }
        }
        if(res) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}