OJ_1044 Pre-Post

#include <iostream>#include <string>using namespace std;long long getcomb(int n,int m){     long long a=1;     long long b=1;     for(int i=0;i<m;i++)     {             a*=n-i;             b*=m-i;     }     return a/b;}long long getresult(string a,string b,int n){            long long sum=1;       if(a.size()==1)                      return sum;       else{            int count=0;            string sub1,sub2;            a=a.substr(1);            b=b.substr(0,b.size()-1);                        while(a.size()!=0)            {                 int pos=b.find(a[0]);                 sub1=a.substr(0,pos+1);                 sub2=b.substr(0,pos+1);                 a=a.substr(pos+1);                 b=b.substr(pos+1);                 count++;                 sum*=getresult(sub1,sub2,n);            }            return sum*getcomb(n,count);       }     }void func(){    int m;    string s1,s2;    while(cin>>m)    {                 if(m==0)break;                 cin>>s1>>s2;                 long long ret=getresult(s1,s2,m);                 cout<<ret<<endl;    }   }int main(int argc, char *argv[]){    //printf("Hello, world\n");func();return 0;}

给出前序后序，求中序情况个数

分析：

例如：

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

同样根节点为a，然后是 bc cb

b,c在两层 C（2,1） * C（2,1）=4

对于这个就有些复杂了,

13 abejkcfghid jkebfghicda

第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3)

再继续递归下去，直到字符串长度为1

题目描述：

        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. 

输入：

        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.

输出：

        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. 

样例输入：

2 abc cba2 abc bca10 abc bca13 abejkcfghid jkebfghicda

样例输出：

4145207352860