九度 oj 题目1044：Pre-Post （确定树的状态）

http://ac.jobdu.com/problem.php?pid=1044

参考了

1.http://blog.youkuaiyun.com/yangnanhai93/article/details/40658571

2.http://www.acmerblog.com/jiudu-1044-2225.html

#include <stdio.h>
#include <cstring>
static int C[22][22];

void getC(){ 
    C[0][1] = C[1][1] = 1;
    for (int i = 2; i <=20; ++i) { 
        C[0][i] = 1;
        for (int j = 1;  j<=i; ++j) { 
            C[j][i] = C[j-1][i-1] + C[j][i-1]; 
        } 

    }  
} 

int solve(char *pre, char* post,int m){ 
    int len = (int) strlen(pre);  
    if(len == 1) return 1; // only a root, status is unique 
    pre = pre + 1;  // remove the root
    post[len-1] = '\0';
    //find sub-tree
    int count = 0; //number of sub-tree
    int res = 1;   //number of tree-status 
    while(*pre){ 
        int i=0;
        char newPre[27],newPost[27]; 
        while(pre[0]!=post[i]){ 
            newPre[i] = pre[i];
            newPost[i] = post[i]; 
            i++;
        }  
        newPre[i] = pre[i];
        newPost[i] = post[i];
        newPre[i+1] = '\0';
        newPost[i+1] = '\0';
        count++; 
        pre = pre+i+1;
        post = post+i+1;
        res = res*solve(newPre,newPost,m); 
    }  
    res = res*C[count][m];
    return res;

}  


int main(){ 
    //freopen("in/1044.in","r",stdin); 
    int m; 
    char pre[27],post[27];
    getC();
    while(scanf("%d %s %s",&m,pre,post) !=EOF){ 
        printf("%d\n",solve(pre,post,m)); 
    }  
}