【洛谷刷题记录|P1219|深度优先搜索|C++】八皇后

大名鼎鼎的八皇后问题就不重复题目了，直接讲思路。

1. 列的方向即深度的方向，沿着列逐层搜索
2. 将4个条件转化为布尔数组，其中“对角线”的条件可以对应平面坐标系中直线的表达式（和或差不变）
3. 如果这层还没搜索过，那就开始搜索；如果这个点可以占用，那就占用，并把不能放棋子的坐标全部标记
4. 别忘了在上一层return之后，把所有标记复原

题解

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
int n;
int row_visited[14];    

int x_is_avaliable[14];   
int y_is_avaliable[14];
int left_cross_is_avaliable[66];     // x+y = ?
int right_cross_is_avaliable[66];     // x-y = ?

int which_column[14];     // result
int solutions = 0;

void init(){
    memset(row_visited, 0, sizeof(row_visited));
    memset(which_column, 0, sizeof(which_column));
    memset(x_is_avaliable, 1, sizeof(x_is_avaliable));
    memset(y_is_avaliable, 1, sizeof(y_is_avaliable));
    memset(left_cross_is_avaliable, 1, sizeof(left_cross_is_avaliable));
    memset(right_cross_is_avaliable, 1, sizeof(right_cross_is_avaliable));
}

void dfs(int row_depth){
    if(row_depth == n+1){
        solutions++;
        // if(solutions <= 3){
            for(int i = 1; i < row_depth; i++){
                cout<<which_column[i]<<" ";
            }
            cout<<endl;
        // }
        return;
    }
    for(int i = 1; i <= n; i++){
        // i is column
        if(row_visited[row_depth] == 0){
            row_visited[row_depth] = 1;
            if(x_is_avaliable[i] && y_is_avaliable[row_depth] && left_cross_is_avaliable[i+row_depth] && right_cross_is_avaliable[i-row_depth+13]){ 
                // i-row+13, prevent negative
                x_is_avaliable[i] = 0;
                y_is_avaliable[row_depth] = 0;
                left_cross_is_avaliable[i+row_depth] = 0;
                right_cross_is_avaliable[i-row_depth+13] = 0;
                which_column[row_depth] = i;
                dfs(row_depth+1);
                which_column[row_depth] = 0;
                x_is_avaliable[i] = 1;
                y_is_avaliable[row_depth] = 1;
                left_cross_is_avaliable[i+row_depth] = 1;
                right_cross_is_avaliable[i-row_depth+13] = 1;
            }   
            row_visited[row_depth] = 0;
        }
    }

}

int main(){
    init();
    cin>>n;
    dfs(1);
    cout<<solutions;
    return 0;
}

示例

输入 6
输出：
2 4 6 1 3 5
3 6 2 5 1 4
4 1 5 2 6 3
5 3 1 6 4 2
4