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 (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​​≤Nfor 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

NO

Sample Input:

[code]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:

[code]YES
NO
NO
YES

[code]#include<iostream>
#include<vector>
#include<unordered_set>

using namespace std;
int main(){

int k;
cin >> k;

for(int i = 0; i < k; i++){

int n;
int graph[1001][1001] = {0};
scanf("%d", &n);
unordered_set<int> un_set;
vector<int> vec(n + 1);
for(int j = 1; j <= n; j++){
int temp;
scanf("%d", &temp);
graph[j][temp] = 1;
// graph[temp][j] = 1;
vec[j] = temp;
un_set.insert(temp);

}

bool flag = true;
for(int j = 1; j <= n; j++){
int p = j + 1;
for(int q = vec[j] + 1; q <= n && p <= n; q++){
if(graph
[q] == 1){
flag = false;
break;
}
p++;
}
if(!flag){
break;
}
p = j - 1;
for(int q = vec[j] - 1; q >= 1 && p>= 1; q--){
if(graph[p][q] == 1){
flag = false;
break;
}
p--;
}
if(!flag){
break;
}

}
if(flag && un_set.size() == n){

printf("YES\n");
}else{

printf("NO\n");
}

}

return 0;
}

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