第六章部分例题

Posted 2020-10-01 lan126

 

没看解答敲了一遍,发现自己题目的理解能力有点差

 

  1 #include <iostream>
  2 #include <cstdio>
  3 
  4 using namespace std;
  5 
  6 struct Node
  7 {
  8     char value;
  9     Node* ch1;
 10     Node* ch2;
 11     Node* ch3;
 12     Node* ch4;
 13 
 14     Node():ch1(NULL),ch2(NULL),ch3(NULL),ch4(NULL){}
 15 };
 16 
 17 Node* newnode() {return new Node();}
 18 
 19 Node* builtree(Node* root)
 20 {
 21     char v;
 22     cin>>v;
 23 
 24     if(root==NULL)
 25         root=newnode();
 26 
 27     if(v==‘e‘)
 28     {
 29         root->value=‘e‘;
 30         return root;
 31     }
 32 
 33     if(v==‘f‘)
 34     {
 35         root->value=‘f‘;
 36         return root;
 37     }
 38     else
 39     {
 40         root->value=‘p‘;
 41 
 42         root->ch1=builtree(root->ch1);
 43         root->ch2=builtree(root->ch2);
 44         root->ch3=builtree(root->ch3);
 45         root->ch4=builtree(root->ch4);
 46     }
 47 
 48     return root;
 49 }
 50 
 51 void _builtree(Node*& root)                //使用引用,不然不能改变root
 52 {
 53     char v;
 54     cin>>v;
 55 
 56     if(root==NULL)
 57         root=newnode();
 58 
 59     if(v==‘e‘)
 60     {
 61         root->value=‘e‘;
 62         return;
 63     }
 64 
 65     if(v==‘f‘)
 66     {
 67         root->value=‘f‘;
 68         return;
 69     }
 70     else
 71     {
 72         root->value=‘p‘;
 73 
 74         _builtree(root->ch1);
 75         _builtree(root->ch2);
 76         _builtree(root->ch3);
 77         _builtree(root->ch4);
 78     }
 79 }
 80 
 81 void reset(Node* root) {root->ch1=root->ch2=root->ch3=root->ch4=NULL;}
 82 
 83 Node* merge(Node* root1,Node* root2)
 84 {
 85     if(root2->value==‘f‘)
 86     {
 87         reset(root1);
 88         root1->value=‘f‘;
 89     }
 90     if(root2->value==‘p‘)
 91     {
 92         if(root1->value==‘e‘) root1=root2;
 93         if(root1->value==‘p‘)
 94         {
 95             root1->ch1=merge(root1->ch1,root2->ch1);
 96             root1->ch2=merge(root1->ch2,root2->ch2);
 97             root1->ch3=merge(root1->ch3,root2->ch3);
 98             root1->ch4=merge(root1->ch4,root2->ch4);
 99         }
100     }
101 
102     return root1;
103 }
104 
105 
106 
107 void print_tree(Node* root)
108 {
109     if(root==NULL) return;
110 
111     printf("%c ",root->value);
112 
113     print_tree(root->ch1);
114     print_tree(root->ch2);
115     print_tree(root->ch3);
116     print_tree(root->ch4);
117 }
118 
119 
120 
121 int main()
122 {
123     Node* root1=NULL;
124     Node* root2=NULL;
125 
126     _builtree(root1);
127     _builtree(root2);
128 
129     root1=merge(root1,root2);
130 
131     print_tree(root1);
132     cout<<endl;
133 
134     return 0;
135 }

虽然能实现四叉树的合并但并不能算出像素:-(

 

然后看答案后又敲了遍

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 const int len=32;
 8 const int maxn=10000;
 9 int w[len][len];
10 char buf[maxn];
11 int cnt;
12 
13 
14 void draw(int r,int c,int length,int& p)
15 {
16     char ch=buf[p++];
17 
18     if(ch==‘p‘)
19     {
20         draw(r+length/2,c,length/2,p);
21         draw(r,c,length/2,p);
22         draw(r,c+length/2,length/2,p);
23         draw(r+length/2,c+length/2,length/2,p);
24     }
25     if(ch==‘f‘)                             //当树满足条件是,相当于递归基
26     {
27         for(int i=c;i<c+length;i++)
28             for(int j=r;j<r+length;j++)
29             {
30                 if(w[i][j]==0)
31                 {
32                     w[i][j]=1;
33                     cnt++;
34                 }
35             }
36     }
37 }
38 
39 int main()
40 {
41     int T;
42     cin>>T;
43 
44     while(T--)
45     {
46         memset(w,0,sizeof(w));
47 
48         cnt=0;
49         int r=0;
50         int c=0;
51         int p=0;
52 
53         scanf("%s",buf);
54         draw(r,c,len,p);
55 
56         scanf("%s",buf);
57         p=0;
58         draw(r,c,len,p);
59 
60         printf("%d\n", cnt);
61     }
62 }