A. Train and Peter

题目连接：

http://www.codeforces.com/contest/8/problem/A

Description

Peter likes to travel by train. He likes it so much that on the train he falls asleep.

Once in summer Peter was going by train from city A to city B, and as usual, was sleeping. Then he woke up, started to look through the window and noticed that every railway station has a flag of a particular colour.

The boy started to memorize the order of the flags’ colours that he had seen. But soon he fell asleep again. Unfortunately, he didn’t sleep long, he woke up and went on memorizing the colours. Then he fell asleep again, and that time he slept till the end of the journey.

At the station he told his parents about what he was doing, and wrote two sequences of the colours that he had seen before and after his sleep, respectively.

Peter’s parents know that their son likes to fantasize. They give you the list of the flags’ colours at the stations that the train passes sequentially on the way from A to B, and ask you to find out if Peter could see those sequences on the way from A to B, or from B to A. Remember, please, that Peter had two periods of wakefulness.

Peter’s parents put lowercase Latin letters for colours. The same letter stands for the same colour, different letters — for different colours.

Input

The input data contains three lines. The first line contains a non-empty string, whose length does not exceed 105, the string consists of lowercase Latin letters — the flags’ colours at the stations on the way from A to B. On the way from B to A the train passes the same stations, but in reverse order.

The second line contains the sequence, written by Peter during the first period of wakefulness. The third line contains the sequence, written during the second period of wakefulness. Both sequences are non-empty, consist of lowercase Latin letters, and the length of each does not exceed 100 letters. Each of the sequences is written in chronological order.

Output

Output one of the four words without inverted commas:

«forward» — if Peter could see such sequences only on the way from A to B;
«backward» — if Peter could see such sequences on the way from B to A;
«both» — if Peter could see such sequences both on the way from A to B, and on the way from B to A;
«fantasy» — if Peter could not see such sequences.

Sample Input

atob
a
b

Sample Output

forward

Hint

题意

有一个人在坐火车，不知道从A到B的还是从B到A的

现在他迷迷糊糊睡了一觉，然后醒来，写下来他看到的字符，然后又迷迷糊糊睡了下去，然后又写下他看到的字符

现在给你这个从A到B的字符串，问这个人坐的火车从A到B还是从B到A的

题解：

首先正着做，暴力求KMP，求匹配的位置，如果存在s1串和s2串匹配的位置不重合的话，就说明合法

正着做一次，再反着做一次就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
char s[maxn],s1[maxn],s2[maxn];
int p[maxn];
vector<int>p1;
vector<int>p2;
int main()
{
    scanf("%s",s+1);
    scanf("%s",s1+1);
    scanf("%s",s2+1);
    int len=strlen(s1+1);
    int j=0;
    for(int i=2;i<=len;i++)
    {
        while(j>0&&s1[j+1]!=s1[i])j=p[j];
        if(s1[j+1]==s1[i])j++;p[i]=j;
    }
    int len2=strlen(s+1);
    j=0;
    for(int i=1;i<=len2;i++)
    {
        while(j>0&&s1[j+1]!=s[i])j=p[j];
        if(s1[j+1]==s[i])j++;
        if(j==len){p1.push_back(i),j=p[j];}
    }
    len = strlen(s2+1);
    j=0;
    memset(p,0,sizeof(p));
    for(int i=2;i<=len;i++)
    {
        while(j>0&&s2[j+1]!=s2[i])j=p[j];
        if(s2[j+1]==s2[i])j++;p[i]=j;
    }
    j=0;
    for(int i=1;i<=len2;i++)
    {
        while(j>0&&s2[j+1]!=s[i])j=p[j];
        if(s2[j+1]==s[i])j++;
        if(j==len){p2.push_back(i),j=p[j];}
    }
    sort(p1.begin(),p1.end());
    sort(p2.begin(),p2.end());
    int flag1 = 0,flag2 = 0;
    if(p1.size()!=0&&p2.size()!=0)
        if(p1[0]+strlen(s2+1)<=p2[p2.size()-1])flag1=1;

    p1.clear(),p2.clear();
    reverse(s+1,s+1+len2);

    memset(p,0,sizeof(p));
    len=strlen(s1+1);
    j=0;
    for(int i=2;i<=len;i++)
    {
        while(j>0&&s1[j+1]!=s1[i])j=p[j];
        if(s1[j+1]==s1[i])j++;p[i]=j;
    }
    len2=strlen(s+1);
    j=0;
    for(int i=1;i<=len2;i++)
    {
        while(j>0&&s1[j+1]!=s[i])j=p[j];
        if(s1[j+1]==s[i])j++;
        if(j==len){p1.push_back(i),j=p[j];}
    }
    len = strlen(s2+1);
    j=0;
    memset(p,0,sizeof(p));
    for(int i=2;i<=len;i++)
    {
        while(j>0&&s2[j+1]!=s2[i])j=p[j];
        if(s2[j+1]==s2[i])j++;p[i]=j;
    }
    j=0;
    for(int i=1;i<=len2;i++)
    {
        while(j>0&&s2[j+1]!=s[i])j=p[j];
        if(s2[j+1]==s[i])j++;
        if(j==len){p2.push_back(i),j=p[j];}
    }
    sort(p1.begin(),p1.end());
    sort(p2.begin(),p2.end());
    if(p1.size()!=0&&p2.size()!=0)
        if(p1[0]+strlen(s2+1)<=p2[p2.size()-1])flag2=1;

    if(flag1&&flag2)
        return puts("both"),0;
    if(flag1)
        return puts("forward"),0;
    if(flag2)
        return puts("backward"),0;
    return puts("fantasy"),0;

}