第六章部分例题

Posted 2020-09-30 lan126

 1 #include <iostream>
 2 #include <sstream>
 3 
 4 using namespace std;
 5 
 6 const int maxn=100000;
 7 int n;
 8 int postorder[maxn],inorder[maxn];
 9 int lh[maxn],rh[maxn];
10 
11 
12 bool read_list(int* a)
13 {
14     string line;
15 
16     if(!getline(cin,line)) return false;
17                                 /*利用stringstream来构造字符串流读入一行数据*/
18     stringstream ss(line);
19 
20     n=0;
21     int num;
22     while(ss>>num) a[n++]=num;
23 
24     return n>0;
25 }
26 
27 int built_tree(int l1,int r1,int l2,int r2)
28 {                              /*递归建立二叉树返回数根,用各个节点的值来标志*/
29     if(l1>r1) return 0;
30 
31     int root=postorder[r2];
32 
33     int p=l1;
34 
35     while(inorder[p]!=root) p++;
36 
37     int cnt=p-l1;
38 
39     lh[root]=built_tree(l1,p-1,l2,l2+cnt-1);
40     rh[root]=built_tree(p+1,r1,l2+cnt,r2-1);
41 
42     return root;
43 }
44 
45 int best_sum,best;
46 
47 void dfs(int u,int sum)
48 {                             /*深度优先,如果是叶子节点就进行判断*/
49     sum+=u;
50 
51     if(!lh[u] && !rh[u])
52     {
53         if(sum<best_sum || (sum==best_sum && u<best))
54         {
55             best_sum=sum;
56             best=u;
57         }
58     }
59 
60     if(lh[u]) dfs(lh[u],sum);
61     if(rh[u]) dfs(rh[u],sum);
62 }
63 
64 
65 int main()
66 {
67 
68     while(read_list(inorder))
69     {
70         read_list(postorder);
71         built_tree(0,n-1,0,n-1);
72 
73         int u=0,sum=0;
74 
75         best=u;
76         best_sum=100000000;
77 
78         dfs(postorder[n-1],sum);
79 
80         cout<<best<<endl;
81     }
82 
83     return 0;
84 }