Codeforces 25E Test 【Hash】

Codeforces 25E Test

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E. Test

Sometimes it is hard to prepare tests for programming problems. Now Bob is preparing tests to new problem about strings — input data to his problem is one string. Bob has 3 wrong solutions to this problem. The first gives the wrong answer if the input data contains the substring s1, the second enters an infinite loop if the input data contains the substring s2, and the third requires too much memory if the input data contains the substring s3. Bob wants these solutions to fail single test. What is the minimal length of test, which couldn’t be passed by all three Bob’s solutions?

Input

There are exactly 3 lines in the input data. The i-th line contains string si. All the strings are non-empty, consists of lowercase Latin letters, the length of each string doesn’t exceed 105.

Output

Output one number — what is minimal length of the string, containing s1, s2 and s3 as substrings.

Examples

input

ab

bc

cd

output

4

input

abacaba

abaaba

x

output

11

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hash题，恶心了一个小时

我的奇葩思路：预处理a和b数组，分别是从前到后和从后往前的hash值前缀和

然后检查一下有没有包含关系

做完了细节比较多

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#include<bits/stdc++.h>
using namespace std;
#define N 100010
#define LL long long
string s[3];
LL a[3][N],b[3][N],ans=0x3f3f3f3f;
LL base=131,Mod=100000007;
LL bas[N],vis[3]={0};
void solve(LL id){
    LL len=s[id].size();
    for(LL i=0;i<len;i++)a[id][i]=((i>0)*a[id][i-1]+s[id][i]*bas[i])%Mod;
    for(LL i=len-1;i>=0;i--)b[id][i]=(b[id][i+1]+s[id][i]*bas[i])%Mod;
}
bool check_in(LL id1,LL id2){
    LL len1=s[id1].size(),len2=s[id2].size();
    LL h1=a[id1][len1-1],h2=a[id2][len1-1];
    if(h1==h2)return 1;
    for(LL i=len1;i<len2;i++){
        h2=(a[id2][i]-a[id2][i-len1]+Mod)%Mod;
        h1=h1*base%Mod;
        if(h1==h2)return 1;
    }
    return 0;
}
LL getlen(LL id1,LL id2){
    LL ans=0,len1=s[id1].size(),len2=s[id2].size();
    for(LL i=1;i<min(len1,len2);i++)
        if(b[id1][len1-i]==(a[id2][i-1]*bas[len1-i]%Mod))ans=max(ans,i);
    return ans;
}
LL work(LL id1,LL id2,LL id3){
    if(vis[id1]){
        if(vis[id2])return s[id3].size();
        if(vis[id3])return s[id2].size();
        return s[id2].size()+s[id3].size()-getlen(id2,id3);
    }
    if(vis[id2]){
        if(vis[id1])return s[id3].size();
        if(vis[id3])return s[id1].size();
        return s[id1].size()+s[id3].size()-getlen(id1,id3);
    }
    if(vis[id3]){
        if(vis[id1])return s[id2].size();
        if(vis[id2])return s[id1].size();
        return s[id1].size()+s[id2].size()-getlen(id1,id2);
    }
    return s[id1].size()+s[id2].size()+s[id3].size()-getlen(id1,id2)-getlen(id2,id3);
}
int main(){
    cin>>s[0]>>s[1]>>s[2];
    if(s[1].size()<s[0].size())swap(s[0],s[1]);
    if(s[2].size()<s[1].size())swap(s[1],s[2]);
    if(s[1].size()<s[0].size())swap(s[0],s[1]);
    LL maxlen=s[2].size();
    bas[0]=1;
    for(LL i=1;i<=maxlen;i++)bas[i]=(bas[i-1]*base)%Mod;
    for(LL i=0;i<3;i++)solve(i);
    if(check_in(0,1))vis[0]=1;
    if(check_in(0,2))vis[0]=1;
    if(check_in(1,2))vis[1]=1;
    for(LL x=0;x<3;x++)
        for(LL y=0;y<3;y++)
            for(LL z=0;z<3;z++)
                if(x!=y&&y!=z&&x!=z)
                    ans=min(ans,work(x,y,z));
    cout<<ans;
    return 0;
}