A1128. N Queens Puzzle

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

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<queue>
 5 #include<vector>
 6 using namespace std;
 7 int num[1001], N, K, hashTB[1001];
 8 int main(){
 9     scanf("%d", &K);
10     for(int i = 0; i < K; i++){
11         scanf("%d", &N);
12         fill(hashTB, hashTB + 1001, 0);
13         for(int j = 1; j <= N; j++){
14             scanf("%d", &num[j]);
15             hashTB[num[j]]++;
16         }
17         int tag = 1;
18         for(int j = 1; j <= N; j++){
19             if(hashTB[j] != 1){
20                 tag = 0;
21                 break;
22             }
23             int m = j - 1, n = num[j] - 1;
24             while(m >= 1 && m <= N && j >= 1 && j <= N){
25                 if(num[m] == n){
26                     tag = 0;
27                     break;
28                 }
29                 m--; n--;
30             }
31             m = j + 1; n = num[j] + 1;
32             while(m >= 1 && m <= N && j >= 1 && j <= N){
33                 if(num[m] == n){
34                     tag = 0;
35                     break;
36                 }
37                 m++; n++;
38             }
39         }
40         if(tag == 0)
41             printf("NO\n");
42         else printf("YES\n");
43     }
44     cin >> N;
45     return 0;
46 }
View Code

总结:

1、由于已经保证了不在同一列,所以只需要检查行和斜线即可。

2、检查a、b两点间的斜线,可用abs(Xa - Xb) == abs(Ya - Yb)。

posted @ 2018-03-04 22:17  ZHUQW  阅读(171)  评论(0编辑  收藏  举报