PAT甲级 1128. N Queens Puzzle (20)
1128. N Queens Puzzle (20)
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.
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|
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| 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
————————————————————————————————————
题目的意思是给出n列皇后位置,判断是否不会互相攻击
思路:n^2暴力判断
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std; #define LL long long
const int INF = 0x3f3f3f3f; int a[10005];
int main()
{
int T,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
int flag=0;
for(int i=1; i<=n; i++)
for(int j=1; j<i; j++)
{
if(a[i]==a[j]||fabs(a[i]-a[j])==fabs(j-i))
{
flag=1;
break;
}
}
printf("%s\n",flag==1?"NO":"YES");
}
return 0;
}
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