八皇后问题求解

最新推荐文章于 2022-04-29 15:39:16 发布

n3verl4nd

最新推荐文章于 2022-04-29 15:39:16 发布

阅读量196

点赞数

本文介绍了一个使用递归回溯法解决八皇后问题的C++程序实现，详细展示了如何通过判断每一行、列及对角线来避免皇后互相攻击，最终输出所有可行解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

描述在国际象棋棋盘上放置八个皇后，要求每两个皇后之间不能直接吃掉对方。输入无输入。输出按给定顺序和格式输出所有八皇后问题的解（见Sample Output）。样例输入

样例输出

No. 1
1 0 0 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 0 0 1 0 0 0 
0 0 0 0 0 0 0 1 
0 1 0 0 0 0 0 0 
0 0 0 1 0 0 0 0 
0 0 0 0 0 1 0 0 
0 0 1 0 0 0 0 0 
No. 2
1 0 0 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 0 1 0 0 0 0 
0 0 0 0 0 1 0 0 
0 0 0 0 0 0 0 1 
0 1 0 0 0 0 0 0 
0 0 0 0 1 0 0 0 
0 0 1 0 0 0 0 0 
No. 3
1 0 0 0 0 0 0 0 
0 0 0 0 0 1 0 0 
0 0 0 0 0 0 0 1 
0 0 1 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 0 1 0 0 0 0 
0 1 0 0 0 0 0 0 
0 0 0 0 1 0 0 0 
No. 4
1 0 0 0 0 0 0 0 
0 0 0 0 1 0 0 0 
0 0 0 0 0 0 0 1 
0 0 0 0 0 1 0 0 
0 0 1 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 1 0 0 0 0 0 0 
0 0 0 1 0 0 0 0 
No. 5
0 0 0 0 0 1 0 0 
1 0 0 0 0 0 0 0 
0 0 0 0 1 0 0 0 
0 1 0 0 0 0 0 0 
0 0 0 0 0 0 0 1 
0 0 1 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 0 1 0 0 0 0 
No. 6
0 0 0 1 0 0 0 0 
1 0 0 0 0 0 0 0 
0 0 0 0 1 0 0 0 
0 0 0 0 0 0 0 1 
0 1 0 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 1 0 0 0 0 0 
0 0 0 0 0 1 0 0 
No. 7
0 0 0 0 1 0 0 0 
1 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 1 
0 0 0 1 0 0 0 0 
0 1 0 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 1 0 0 0 0 0 
0 0 0 0 0 1 0 0 
No. 8
0 0 1 0 0 0 0 0 
1 0 0 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 0 0 1 0 0 0 
0 0 0 0 0 0 0 1 
0 1 0 0 0 0 0 0 
0 0 0 1 0 0 0 0 
0 0 0 0 0 1 0 0 
No. 9
0 0 0 0 1 0 0 0 
1 0 0 0 0 0 0 0 
0 0 0 1 0 0 0 0 
0 0 0 0 0 1 0 0 
0 0 0 0 0 0 0 1 
0 1 0 0 0 0 0 0 
0 0 0 0 0 0 1 0 
0 0 1 0 0 0 0 0 
...以下省略

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define MAX 15
int n, tot;
int C[MAX];

void dfs(int cur) //每一行
{
	if(cur == n)
	{
		tot++;
		printf("No. %d\n", tot);
		for(int i = 0; i < 8; i++)
		{
			for(int j = 0; j < 8; j++)
			{
				if(i == C[j])
				{
					cout << 1 << " ";
				}
				else
				{
					cout << 0 << " ";
				}
			}
			cout << endl;
		}
	}
	else
	{
		for(int i = 0; i < n; i++)
		{
			bool flag = true;
			C[cur] = i;
			for(int j = 0; j < cur; j++)
			{
				if(C[cur] == C[j] || cur - C[cur] == j - C[j] || cur + C[cur] == j + C[j])//每一列、主对角线、副对角线
				{
					flag = false;
					break;
				}
			}
			if(flag)
			{
				dfs(cur + 1);
			}
		}
	}
}

int main()
{
	tot = 0;
	n = 8;
	dfs(0);
	return 0;
}