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
转载来自: https://blog.csdn.net/Hickey_Chen/article/details/81095025
只能说自己脑子秀逗了 这种东西都想不到......实在是无奈....................
get 新姿势
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4+10;
vector<int>vec;
bool isQueen()
{
int size = vec.size();
// printf("%d\n",vec.size());
for(int i = 0; i < size ; i++)
{
for(int j = i + 1; j < size; j++)
{
if(vec[i] == vec[j]) //肯定不同行 判断是否同列
return false;
int colum = j - i ;
if(vec[j] - colum == vec[i] || vec[j] + colum == vec[i])
return false;
}
}
return true;
}
int main()
{
int k;
scanf("%d",&k);
while(k--)
{
vec.clear();
int num,x;
scanf("%d",&num);
for(int i = 1 ; i <= num; i++)
{
scanf("%d", &x);
vec.push_back(x);
}
if(isQueen())
printf("YES\n");
else
printf("NO\n");
}
return 0;
}