Codeforces427D Match & Catch

http://codeforces.com/problemset/problem/427/D

Police headquarter is monitoring signal on different frequency levels. They have got two suspiciously encoded strings s1 and s2 from two different frequencies as signals. They are suspecting that these two strings are from two different criminals and they are planning to do some evil task.

Now they are trying to find a common substring of minimum length between these two strings. The substring must occur only once in the first string, and also it must occur only once in the second string.

Given two strings s1 and s2 consist of lowercase Latin letters, find the smallest (by length) common substring p of both s1 and s2, where pis a unique substring in s1 and also in s2. See notes for formal definition of substring and uniqueness.

Input

The first line of input contains s1 and the second line contains s2 (1 ≤ |s1|, |s2| ≤ 5000). Both strings consist of lowercase Latin letters.

Output

Print the length of the smallest common unique substring of s1 and s2. If there are no common unique substrings of s1 and s2 print -1.

题意：给定两个串，求长度最短的满足<在两个串中均只出现一次>的子串的长度。

思路：用字符串hash

枚举子串长度l，对于两个串中每个长度为l的子串，分别加到两个map中计数，最后遍历一个map，若某子串在两个串均只出现一次，那么可行。但是这样的复杂度是O(n^2*log(n))，会超时。

把map记录次数改成hash方法，就去掉了log(n)，复杂度O(n^2)，可以过。

其实这题方法特别多，hash最简单了。

#include<bits/stdc++.h>
using namespace std;
#define maxn 5000+100
#define ull unsigned long long

char s1[maxn],s2[maxn];
int len1,len2;
int x=123;
ull H1[maxn],H2[maxn],xp[maxn];

const int hashsize=100007,maxstate=10100;
struct HashMap{
    int head[hashsize];
    int next[maxstate];
    ull state[maxstate];
    int num_s1[maxstate];
    int num_s2[maxstate];
    int sz;

    void init()
    {
        sz=0;
        memset(head,-1,sizeof(head));
    }

    void insert(ull v,int type)
    {
        int h=v%hashsize;
        for(int u=head[h];u!=-1;u=next[u])
        {
            if(state[u]==v)
            {
                if(type==1)num_s1[u]++;
                else num_s2[u]++;
                return;
            }
        }
        num_s1[sz]=0;
        num_s2[sz]=0;
        state[sz]=v;
        next[sz]=head[h];
        if(type==1)num_s1[sz]++;
        else num_s2[sz]++;
        head[h]=sz++;
    }

    bool check()
    {
        for(int i=0;i<sz;i++)if(num_s1[i]==1&&num_s2[i]==1)return true;
        return false;
    }
}H;

void strhash_init()
{
    len1=strlen(s1);
    len2=strlen(s2);
    for(int i=len1-1;i>=0;i--)H1[i]=H1[i+1]*x+s1[i];
    for(int i=len2-1;i>=0;i--)H2[i]=H2[i+1]*x+s2[i];
    xp[0]=1;
    for(int i=1;i<=5000;i++)xp[i]=xp[i-1]*x;
}

ull get_seg(int p,int l,int type)
{
    if(type==1)return H1[p]-H1[p+l]*xp[l];
    else return H2[p]-H2[p+l]*xp[l];
}

int main()
{
    //freopen("input.in","r",stdin);
    scanf("%s%s",s1,s2);
    strhash_init();
    for(int l=1;l<=min(len1,len2);l++)
    {
        H.init();
        for(int i=0;i+l-1<len1;i++)H.insert(get_seg(i,l,1),1);
        for(int i=0;i+l-1<len2;i++)H.insert(get_seg(i,l,2),2);
        if(H.check())
        {
            cout<<l<<endl;
            exit(0);
        }
    }
    cout<<-1<<endl;   
    return 0;
}