UVALive - 5844

题是pdf版

Sample Input
2
3
mississippi
nni55i55ippi
2
foobar
|=o08ar
Sample Output
1
0

/**
    题意：给出一个normal串，一个leet串，看能否符合关系的映射
    做法：dfs 将两个串进行匹配注意初始化
**/
#include <iostream>
#include <string.h>
#include <stdio.h>
#include  <cmath>
#include <algorithm>
#define maxn 110
using namespace std;
char ch[maxn];
char ch1[maxn];
char alp[][];
char c[];
int dfs(int k,int a,int b,int len,int len1)
{
    if(a == len && b == len1) return ;
    if(a == len || b == len1) return ;
    for(int i=; i<k; i++)
    {
        if(b+i<len1)
        {
            memset(c,'\0',sizeof(c));
            for(int j=; j<=i; j++)
            {
                c[j] = ch1[b+j];
            }
            bool isok = false;
            if(strcmp(alp[ch[a]-'a'],"") == )  ///当前字母还没有匹配
            {
                isok = true;
                strcpy(alp[ch[a]-'a'],c);
            }
            if(isok || (strcmp(alp[ch[a]-'a'] ,c) == ))
            {
                if(dfs(k,a+,b+i+,len,len1)) return ;
            }
            if(isok)
            {
                strcpy(alp[ch[a]-'a'],""); ///回溯
            }
        }
    }
    return ;
}
int main()
{
   // freopen("in1.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int k;
        scanf("%d",&k);
        memset(alp,'\0',sizeof(alp));
        scanf("%s %s",ch,ch1);
        int len = strlen(ch);
        int len1 = strlen(ch1);
        int tt = ;

        tt = dfs(k,,,len,len1);
        printf("%d\n",tt);
    }
    return ;
}