D. DZY Loves Strings

题目连接：

http://codeforces.com/contest/444/problem/D

Description

DZY loves strings, and he enjoys collecting them.

In China, many people like to use strings containing their names’ initials, for example: xyz, jcvb, dzy, dyh.

Once DZY found a lucky string s. A lot of pairs of good friends came to DZY when they heard about the news. The first member of the i-th pair has name ai, the second one has name bi. Each pair wondered if there is a substring of the lucky string containing both of their names. If so, they want to find the one with minimum length, which can give them good luck and make their friendship last forever.

Please help DZY for each pair find the minimum length of the substring of s that contains both ai and bi, or point out that such substring doesn’t exist.

A substring of s is a string slsl + 1… sr for some integers l, r (1 ≤ l ≤ r ≤ |s|). The length of such the substring is (r - l + 1).

A string p contains some another string q if there is a substring of p equal to q.

Input

The first line contains a string s (1 ≤ |s| ≤ 50000).

The second line contains a non-negative integer q (0 ≤ q ≤ 100000) — the number of pairs. Each of the next q lines describes a pair, the line contains two space-separated strings ai and bi (1 ≤ |ai|, |bi| ≤ 4).

It is guaranteed that all the strings only consist of lowercase English letters.

Output

For each pair, print a line containing a single integer — the minimum length of the required substring. If there is no such substring, output -1.

Sample Input

xudyhduxyz
3
xyz xyz
dyh xyz
dzy xyz

Sample Output

3
8
-1

题意

给你一个串，然后有q次询问，每次询问给你两个串s1,s2

你需要找到一个最短的串s，使得这个串包含这两个子串

题解：

因为询问的那个串的长度才4嘛

我就预处理所有串的出现的位置，然后每次O(n+m)去找最小的长度就好了

再加一个人尽皆知的剪枝：如果这个询问问过了，那就直接输出答案

然后就莽过去了……

代码

#include <bits/stdc++.h>

using namespace std;

const int N=5e4+10;
const int NN=6e5+10;
char s[N];
vector<int> v[NN];
map<pair<int,int>,int> mp;

int main()
{
    scanf("%s",s);
    int n=strlen(s);
    for(int l=1;l<=4;l++)
    {
        for(int i=0;i+l-1<n;i++)
        {
            int tmp=0;
            for(int j=l-1;j>=0;j--)
                tmp=(tmp*27)+s[i+j]-'a'+1;
            v[tmp].push_back(i);
        }
    }
    int q;
    scanf("%d",&q);
    while(q--)
    {
        char s1[6],s2[6],ss[2];
        scanf("%s%s",s1,s2);
        int t1=0,t2=0,l1=strlen(s1),l2=strlen(s2);
        for(int j=l1-1;j>=0;j--)
            t1=(t1*27)+s1[j]-'a'+1;
        for(int j=l2-1;j>=0;j--)
            t2=(t2*27)+s2[j]-'a'+1;
        if(mp[make_pair(t1,t2)]!=0)
        {
            printf("%d\n",mp[make_pair(t1,t2)]);
            continue;
        }
        int ans=n+2;
        for(int j=0,i=0;i<v[t1].size();i++)
        {
            for(;j<v[t2].size()&&v[t2][j]<v[t1][i];j++);
            if(j>=v[t2].size()) break;
            ans=min(ans,max(v[t2][j]+l2,v[t1][i]+l1)-min(v[t1][i],v[t2][j]));
        }
        for(int j=v[t2].size()-1,i=v[t1].size()-1;i>=0;i--)
        {
            for(;j>=0&&v[t2][j]>v[t1][i];j--);
            if(j<0) break;
            ans=min(ans,max(v[t2][j]+l2,v[t1][i]+l1)-min(v[t1][i],v[t2][j]));
        }
        if(ans==n+2) ans=-1;
        mp[make_pair(t1,t2)]=ans;
        printf("%d\n",ans);
    }
    return 0;
}