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.
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 }
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)。