Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them in the more linear `tree in-order' and `tree pre-order' notations.
Your job is to create the `tree post-order' notation of a cow's heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes.
Here is a graphical representation of the tree used in the sample input and output:
C
/
/
B G
/ /
A D H
/
E F
The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree.
The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree.
The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root.
PROGRAM NAME: heritage
INPUT FORMAT
| Line 1: | The in-order representation of a tree. |
| Line 2: | The pre-order representation of that same tree. |
SAMPLE INPUT (file heritage.in)
ABEDFCHG CBADEFGH
OUTPUT FORMAT
A single line with the post-order representation of the tree.
SAMPLE OUTPUT (file heritage.out)
AEFDBHGC
—————————————————————
用前序遍历中某个字母的位置可以得到它的左子树的长度
然后就可以了
存代码
1 /* 2 ID: ivorysi 3 PROG: heritage 4 LANG: C++ 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <algorithm> 10 #include <queue> 11 #include <set> 12 #include <vector> 13 #include <string.h> 14 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 15 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 16 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 17 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 18 #define inf 0x3f3f3f3f 19 #define MAXN 400005 20 #define ivorysi 21 #define mo 97797977 22 #define ha 974711 23 #define ba 47 24 #define fi first 25 #define se second 26 #define pii pair<int,int> 27 using namespace std; 28 typedef long long ll; 29 int id[30]; 30 int root,lson[30],rson[30],fa[30]; 31 string str; 32 int dfs(string s,int fa) { 33 if(s.length()<1) return 0; 34 int u=s[0]-'A'+1; 35 if(s.length()==1) {return u;} 36 lson[u]=dfs(s.substr(1,id[u]-id[fa]-1),fa); 37 rson[u]=dfs(s.substr(id[u]-id[fa]),u); 38 return u; 39 } 40 void ans(int u) { 41 if(u==0) return; 42 ans(lson[u]); 43 ans(rson[u]); 44 printf("%c",u+'A'-1); 45 } 46 void init() { 47 cin>>str; 48 xiaosiji(i,0,str.length()) { 49 id[str[i]-'A'+1]=i+1; 50 } 51 cin>>str; 52 root=str[0]-'A'+1; 53 } 54 void solve() { 55 init(); 56 dfs(str,0); 57 ans(root); 58 puts(""); 59 } 60 int main(int argc, char const *argv[]) 61 { 62 #ifdef ivorysi 63 freopen("heritage.in","r",stdin); 64 freopen("heritage.out","w",stdout); 65 #else 66 freopen("f1.in","r",stdin); 67 #endif 68 solve(); 69 }