Leetcode: n-queen, n-queen II

思路:

题目给出的测试数据范围比较小, 使用回溯就可以AC, 搞的我也没有兴趣去研究高效解法了

总结:

刚开始, 本以为用棋盘问题的状态压缩 DP 就可以解决, 但做完 N-queen 才发现多个皇后并不能在同一条斜线上, 状态压缩的解法似乎就不好用了

代码:

#include <iostream>
#include <vector>
#include <string>
using namespace std;
class Solution {
public:
	int N;
	vector<int> pos;
	int result;
	bool visited[10];

	bool attack(const int &row, const int &col) {
		for(int i = 0; i < pos.size(); i ++) {
			if(abs(pos[i]-col) == abs(i-row))
				return true;
		}
		return false;
	}
	void dfs(const int &depth) {
		if( depth == N) {
			result += 1;
		}
		for(int i = 0; i < N; i ++) {
			if(visited[i] == 0 ) {
				if( pos.size() > 0 && attack(depth, i))
					continue;
				visited[i] = 1;
				pos.push_back(i);
				dfs(depth+1);
				pos.pop_back();
				visited[i] = 0;
			}
			
		}
	}
    int totalNQueens(int n) {
		pos.clear();
		result=0;
		memset(visited, 0, sizeof(visited));
        N = n;
		dfs(0);
		return result;
    }
};
int main() {
	Solution solution;
	int res = solution.totalNQueens(9);
	cout << res << endl;
	return 0;
}

#include <iostream>
#include <vector>
#include <string>
using namespace std;
class Solution {
public:
	int N;
	vector<int> pos;
	vector<vector<string> > result;
	bool visited[10];

	bool attack(const int &row, const int &col) {
		for(int i = 0; i < pos.size(); i ++) {
			if(abs(pos[i]-col) == abs(i-row))
				return true;
		}
		return false;
	}
	void dfs(const int &depth) {
		if( depth == N) {
			vector<string> temp;
			string str;
			for(int i = 0; i < N; i ++) {
				str.clear();
				for(int j = 0; j < N; j ++) {
					if(pos[i] == j)
						str.push_back('Q');
					else
						str.push_back('.');
				}
				temp.push_back(str);
			}
			result.push_back(temp);		
		}
		for(int i = 0; i < N; i ++) {
			if(visited[i] == 0 ) {
				if( pos.size() > 0 && attack(depth, i))
					continue;
				visited[i] = 1;
				pos.push_back(i);
				dfs(depth+1);
				pos.pop_back();
				visited[i] = 0;
			}
			
		}
	}
    vector<vector<string> > solveNQueens(int n) {
		pos.clear();
		result.clear();
		memset(visited, 0, sizeof(visited));
        N = n;
		dfs(0);
		return result;
    }
};