八皇后问题 OpenJ_Bailian - 2698 DFS

Posted 2020-10-06 小小超plus

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。Input无输入。Output按给定顺序和格式输出所有八皇后问题的解（见Sample Output）。Sample Input

Sample Output

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

Hint此题可使用函数递归调用的方法求解。 

经典的DFS入门题 核心代码就是一个for循环 代码简单易懂  ，但是不知怎么优化 才能让时间变成0。

 1 #include <iostream>
 2 using namespace std;
 3 #include<string.h>
 4 #include<set>
 5 #include<stdio.h>
 6 #include<math.h>
 7 #include<queue>
 8 #include<map>
 9 #include<algorithm>
10 #include<cstdio>
11 #include<cmath>
12 #include<cstring>
13 #include <cstdio>
14 #include <cstdlib>
15 #include<stack>
16 int add=0;
17 int a[20][20];
18 
19 int panduan(int x,int y)
20 {
21     for(int i=0;i<x;i++)
22     {
23         if(a[i][y]==1)
24             return 0;
25     }
26     int y1=y+x;
27     int x1=0;
28     for(;x1<=8&&y1>0;x1++,y1--)
29     {
30         if(a[x1][y1]==1)
31             return 0;
32     }
33     x+=8;
34     y+=8;
35     for(;x>0&&y>0;x--,y--)
36     {
37         if(a[x][y]==1)
38             return 0;
39     }
40         return 1;
41 }
42 
43 
44 void dfs(int x)
45 {
46     if(x==9)
47     {
48         printf("No. %d\\n",++add);
49         for(int i=1;i<=8;i++)
50         {
51             int flag=0;
52             for(int j=1;j<=8;j++)
53             {
54                 if(!flag)
55                 {
56                     flag=1;
57                     printf("%d",a[j][i]);
58                 }
59                 else
60                 {
61                     printf(" %d",a[j][i]);
62                 }
63             }
64             printf("\\n");
65         }
66         return ;
67     }
68     for(int i=1;i<=8;i++)
69     {
70         if(panduan(x,i))
71         {
72             a[x][i]=1;
73             dfs(x+1);
74             a[x][i]=0;
75         }
76     }
77     return ;
78 }
79 
80 int main()
81 {
82     memset(a,0,sizeof(a));
83         dfs(1);
84     return 0;
85 }

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