POJ1240

Pre-Post-erous!

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 2374 Accepted: 1464

Description

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: 

    a   a   a   a   /   /     \   \  b   b       b   b /     \     /     \c       c   c       c

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

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. 

Output

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. 

Sample Input

2 abc cba2 abc bca10 abc bca13 abejkcfghid jkebfghicda0

Sample Output

4145
207352860
这个题也比较简单，写的开心
#include <iostream>#include <cstring>#include <stdio.h>using namespace std;int m,sum;char s1[25],s2[25];int fd(int lpre,int rpre,int lpost,int rpost){    if(lpre==rpre)        return 1;    int cou,nr[20],nl[20],i,j,k,ve,to;    k=1;    i=lpre+1;    j=lpost;    cou=0;    nr[0]=j-1;    nl[0]=i-1;    while(i<=rpre)    {        char tem=s1[i];        //cout<<i<<endl;        for(;j<=rpost-1;j++)        {            i++;            if(tem==s2[j])            {                cou++;                nl[k]=i-1;                nr[k]=j;                k++;                j++;                //cout<<"sdsa"<<endl;                break;            }        }    }    ve=1;    for(i=0;i<cou;i++)        ve=ve*(m-i)/(i+1);    //cout<<cou<<endl<<ve<<endl;    to=1;    for(i=0;i<cou;i++){        //cout<<nl[i]+1<<"-"<<nl[i+1]<<endl;        //cout<<nr[i]+1<<"-"<<nr[i+1]<<endl;        to=to*fd(nl[i]+1,nl[i+1],nr[i]+1,nr[i+1]);    }    return to*ve;}int main(){    cin>>m>>s1>>s2;    int lpre,rpre,lpost,rpost,len;    while(m!=0)    {        len=strlen(s1);        lpre=0;rpre=len-1;        lpost=0;rpost=len-1;        cout<<fd(lpre,rpre,lpost,rpost)<<endl;        cin>>m>>s1>>s2;    }    return 0;}