Skip Letter Code

Description

Romeo and Juliet are college students in the city of Verona and are in love with each other. Unfortunately, both their families are not very glad with this fact. They are trying to prevent any attempts of young people to communicate. Thus, Romeo and Juliet had to introduce their special code to be capable of writing love letters to each other. Both of them have a little book of Shakespeare sonnets and agreed to use this book as a dictionary for their code. The code is like follows: only words that appear in the dictionary may be used in the love letter. Then, the person who writes the letter may skip some letters in each word. Of course that person tries to skip as may letters as possible, but he would keep in mind that the letter should be decoded with just one possibility for each word. Oh, such a difficult task! Fortunately, both of the lovers have a personal computer and you can (can't you?) help them writing the program, that being given with the dictionary and the coded message decodes one if there is just one way to decode the message or reports that there could be more than one decoding possibility. Thus, you will help both writing and reading persons.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

First line of the input contains single integer 1 <= N <= 100 - number of words in a dictionary. Next N lines each contain one word from the dictionary. Each word in a dictionary is not longer than 5 characters and (as well as coded message) consists from capital letters A to Z. The rest of file (up to # character) contains coded message (no longer than 10^6 characters) which is coded words separated by space(s) and/or newline characters.

Output

Output should contain single line with word AMBIGUITY if there could be more than one possible decoding of the message or decoded message with all spaces and newline characters kept intact. Note, that # character should not be printed.

Sample Input

2

3
I
LOVE
YOU
I E
U#

3
I
LOVE
YOU
I O
Y#

Sample Output

I LOVE
YOU

AMBIGUITY

这道题是需要你将一个连续的字符串进行解码，注意是连续的，而且是有顺序的，如果对应编码中有

多个答案那么就输出AMBIGUITY，所以需要映射是唯一的才行，可以用组合将编码的所有可能列举出

来，然后采用map存储位置，这样后面就只需要查询了。

#include <stdio.h>#include <string.h>#include <string>#include <map>#include <iostream>using namespace std;map < string, int > vis;const int N = 105, maxn = 1000005;char word[N][N], op[N], ans[maxn];int s, len;void dfs ( int k, string str, int n, int pos ){    if ( k > n || len-pos < n-k )   //剪枝        return ;    if ( k == n )    {        if ( vis[str] == 0 || vis[str] == s )        //如果里面没数或者相等就赋值            vis[str] = s;        else    //否则变成-1表示有多种可能            vis[str] = -1;        return ;    }    for ( int i = pos; i < len; i ++ )  //组合数要从i+1开始        dfs ( k+1, str+word[s][i], n, i+1 );}int is_letter ( char ch ){    return ch >= 'A' && ch <= 'Z' || ch >= 'a' && ch <= 'z';}int main ( ){    int T, n, tag, cnt;    //freopen ( "in0.in", "r", stdin );    scanf ( "%d", &T );    string str;    while ( T -- )    {        tag = 0;        vis.clear ( );        scanf ( "%d", &n );        for ( int i = 1; i <= n; i ++ )            scanf ( "%s", word[i] );        for ( int i = 1; i <= n; i ++ )        {            str = "";            len = strlen ( word[i] );            for ( int j = 1; j <= len; j ++ )   //使用编码的长度            {                s = i;                dfs ( 0, str, j, 0 );            }        }        getchar ( );        strcpy ( ans, "" );        cnt = 0;        while ( gets ( op ) )        {            int size = strlen ( op );            if ( tag )            {                if ( op[size-1] == '#' )                    break ;                continue ;            }            for ( int i = 0; i < size; i ++ )            {                string ch = "";                if ( is_letter ( op[i] ) )                {                    while ( is_letter ( op[i] ) )                    {                        ch = ch+op[i];                        i ++;                    }                    i --;                    if ( vis[ch] == -1 || vis[ch] == 0 )                        tag = 1;                    else                    {   int pos = vis[ch];                        int l = strlen ( word[pos] );                        for ( int i = 0; i < l; i ++ )                            ans[cnt ++] = word[pos][i];                    }                }                else if ( op[i] != '#' )                    ans[cnt ++] = op[i];            }            ans[cnt ++] = '\n'; //注意在后面加换行            ans[cnt] = '\0';            if ( op[size-1] == '#' )                break ;        }        printf ( "%s", tag ? "AMBIGUITY\n" : ans );        if ( T )    //两组数据中有一个换行            printf ( "\n" );    }    return 0;}