2014
04-02

# 九度-1044-Pre-Post[分治和组合数]

We are all familiar with pre-order, in-order and post-order traversals of binary trees. A common problem in data structure classes is to find the pre-order traversal of a binary tree when given the in-order and post-order traversals. Alternatively, you can find the post-order traversal when given the in-order and pre-order. However, in general you cannot determine the in-order traversal of a tree when given its pre-order and post-order traversals. Consider the four binary trees below:

All of these trees have the same pre-order and post-order traversals. This phenomenon is not restricted to binary trees, but holds for general m-ary trees as well.

Input will consist of multiple problem instances. Each instance will consist of a line of the form
m s1 s2
indicating that the trees are m-ary trees, s1 is the pre-order traversal and s2 is the post-order traversal.All traversal strings will consist of lowercase alphabetic characters. For all input instances, 1 <= m <= 20 and the length of s1 and s2 will be between 1 and 26 inclusive. If the length of s1 is k (which is the same as the length of s2, of course), the first k letters of the alphabet will be used in the strings. An input line of 0 will terminate the input.

For each problem instance, you should output one line containing the number of possible trees which would result in the pre-order and post-order traversals for the instance. All output values will be within the range of a 32-bit signed integer. For each problem instance, you are guaranteed that there is at least one tree with the given pre-order and post-order traversals.

2 abc cba
2 abc bca
10 abc bca
13 abejkcfghid jkebfghicda

4
1
45
207352860

10 abc bca
根节点为a是确定的，接下来是 bc bc
可知b,c在同一级别，有C（10,2）=45 (10个位置中取两个)
2 abc cba

b,c在两层 C（2,1） * C（2,1）=4

13 abejkcfghid jkebfghicda

#include <iostream>
#include <string.h>
#include <string>
#include <stdio.h>
using namespace std;

char str1[30],str2[30];
string pre,post;
int n;
int c[23][23];
//组合数计算
void initc(){
c[1][0] = c[1][1] = 1;
for(int i=2; i<23; i++){
c[i][0] = 1;
for(int j=1; j<=i; j++){
c[i][j] = c[i-1][j] + c[i-1][j-1];
}
}
}
int getCnt(string left,string right){
int cnt=0;
int res = 1;
if(left.size() <= 1) return n; //只有一个元素，可以放置n个位置
for(int i=0; i<left.length();){
int j=i;
while(j<left.length() && right[j] != left[i]) j++ ;
//一层的单个叶子节点不考虑
if(j>i)
res *= getCnt(left.substr(i+1,j-i), right.substr(i,j-i));
cnt++;
i = j+1;
}
res *= c[n][cnt];
return res;
}

int main() {
freopen("in.txt","r",stdin);
initc();
while(cin >> n ){
if(!n) break;
cin >>  pre >> post;
int ans = 1;
if(pre.length() > 1)
ans = getCnt(pre.substr(1,pre.length()-1), post.substr(0, pre.length()-1) );
cout << ans << endl;
}
return 0;
}

#include <stdio.h>
#include <string>
#include <iostream>
using namespace std;
int c[21][21];
int n;
long long test(string pre, string post) {
long long sum = 1;
int num = 0;
int k = 0, i;
pre.erase(pre.begin());
post=post.substr(0, post.length()-1);
while (k < pre.length()) {
for (i = 0; i < post.length(); i++)
if (pre[k] == post[i]) {
sum *= test(pre.substr(k, i - k + 1),
post.substr(k, i - k + 1));
num++; //num代表串被分成了几段(例如 (bejkcfghid,jkebfghicd) = (bejk, cfghi, d) )
k = i + 1;
break;
}
}
//cout << pre << "  " << post <<"  " << t1 << " =" << num << endl << endl;
sum *= c[num][n]; //从n中取num个的取法个数
return sum;
}
void getsc() {
int i, j;
c[0][1] = c[1][1] = 1;
for (i = 2; i < 21; i++) {
c[0][i] = 1;
for (j = 1; j <= i; j++){
if (i == j)
c[j][i] = 1;
else
c[j][i] = c[j - 1][i - 1] + c[j][i - 1];

}
}
}
int main() {
string pre, post;
getsc();
while ((cin >> n >> pre >> post) && n) {

cout << test(pre, post) << endl; //printf("%ld\n",test(pre,post));
}
return 0;
}
/**************************************************************
Problem: 1044
User: coder
Language: C++
Result: Accepted
Time:0 ms
Memory:1520 kb
****************************************************************/

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2. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？

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4. 代码是给出了，但是解析的也太不清晰了吧！如 13 abejkcfghid jkebfghicda
第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3)，为什么要这样拆分，原则是什么？

• a是根先忽略掉，递归子树。剩下前缀bejkcfghid和后缀jkebfghicd，分拆的原则的是每个子树前缀和后缀的节点个数是一样的，根节点出现在前缀的第一个，后缀的最后一个。根节点b出现后缀的第四个位置，则第一部分为四个节点，前缀bejk，后缀jkeb，剩下的c出现在后缀的倒数第2个，就划分为cfghi和 fghic，第3部分就为c、c