1128 N Queens Puzzle (20分)

最新推荐文章于 2021-09-08 19:55:50 发布

zbchenchanghao

最新推荐文章于 2021-09-08 19:55:50 发布

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本文介绍了一个简化版的八皇后问题，即不需求解而仅判断给定的棋盘配置是否为有效解。通过输入一系列棋盘配置，利用整数序列表示各列皇后的位置，程序检查这些配置是否满足八皇后问题的条件，即任意两个皇后不在同一行、列或对角线上。

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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≤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 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

#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[p][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;
}