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棋盘覆盖问题

时间:2015-09-15 18:11:35      阅读:173      评论:0      收藏:0      [点我收藏+]

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 1 #include<iostream>
 2 using namespace std;
 3 #define MAX 1024 
 4 int a[MAX][MAX];
 5 int num=1;
 6 void solve(int fromx,int fromy,int x,int y,int size)
 7 {
 8     if(size==1)
 9         return;
10     if(x<fromx+size/2&&y<fromy+size/2)
11     {
12         a[fromx+size/2][fromy+size/2]=num;
13         a[fromx+size/2-1][fromy+size/2]=num;
14         a[fromx+size/2][fromy+size/2-1]=num++;
15         solve(fromx,fromy,x,y,size/2);
16         solve(fromx,fromy+size/2,fromx+size/2-1,fromy+size/2,size/2);
17         solve(fromx+size/2,fromy,fromx+size/2,fromy+size/2-1,size/2);
18         solve(fromx+size/2,fromy+size/2,fromx+size/2,fromy+size/2,size/2);
19     }
20     if(x<fromx+size/2&&y>=fromy+size/2)
21     {
22         a[fromx+size/2][fromy+size/2]=num;
23         a[fromx+size/2-1][fromy+size/2-1]=num;
24         a[fromx+size/2][fromy+size/2-1]=num++;
25         solve(fromx+size/2,fromy,x,y,size/2);
26         solve(fromx,fromy+size/2,fromx+size/2-1,fromy+size/2,size/2);
27         solve(fromx,fromy,fromx+size/2-1,fromy+size/2-1,size/2);
28         solve(fromx+size/2,fromy+size/2,fromx+size/2,fromy+size/2,size/2);
29     }
30     if(x>=fromx+size/2&&y<fromy+size/2)
31     {
32         a[fromx+size/2][fromy+size/2]=num;
33         a[fromx+size/2-1][fromy+size/2]=num;
34         a[fromx+size/2-1][fromy+size/2-1]=num++;
35         solve(fromx+size/2,fromy,x,y,size/2);
36         solve(fromx,fromy+size/2,fromx+size/2-1,fromy+size/2,size/2);
37         solve(fromx,fromy,fromx+size/2-1,fromy+size/2-1,size/2);
38         solve(fromx+size/2,fromy+size/2,fromx+size/2,fromy+size/2,size/2);
39     }
40     if(x>=fromx+size/2&&y>=fromy+size/2)
41     {
42         a[fromx+size/2-1][fromy+size/2-1]=num;
43         a[fromx+size/2-1][fromy+size/2]=num;
44         a[fromx+size/2][fromy+size/2-1]=num++;
45         solve(fromx+size/2,fromy+size/2,x,y,size/2);
46         solve(fromx,fromy+size/2,fromx+size/2-1,fromy+size/2,size/2);
47         solve(fromx+size/2,fromy,fromx+size/2,fromy+size/2-1,size/2);
48         solve(fromx,fromy,fromx+size/2-1,fromy+size/2-1,size/2);
49     }
50 
51 }
52 int main()
53 {
54     int k,x,y,i,j;
55     while(cin>>k)
56     {
57         cin>>x>>y;
58         int size=1<<k;
59         a[x][y]=0;
60         solve(0,0,x,y,size);
61         for(i=0;i<size;i++)
62         {
63             for(j=0;j<size;j++)
64                 cout<<a[i][j]<<"    ";
65             cout<<endl;
66         }
67     }
68     return 0;
69 }

输入:对每个测试有两行,第一行是k,第二行是特殊方格位置坐标x,y。

输出:边长为2的k次幂的方阵,特殊方格编号为0。

eg.

输入:

  2

  0  1

输出:

  2 0 3 3
  2 2 1 3
  4 1 1 5
  4 4 5 5

棋盘覆盖问题

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原文地址:http://www.cnblogs.com/xdbingo/p/4810818.html

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