POJ 1226 Substrings(后缀数组)

题目链接：http://poj.org/problem?id=1226

题意：给定n个串。求一个最长的串，使得这个串或者其反串在每个串中都出现过？

思路：二分。

int r[N],sa[N],wa[N],wb[N],wd[N],rank[N],h[N];



int cmp(int *r,int a,int b,int len)

{

    return r[a]==r[b]&&r[a+len]==r[b+len];

}



void da(int *r,int *sa,int n,int m)

{

    int i,j,p,*x=wa,*y=wb,*t;

    FOR0(i,m) wd[i]=0;

    FOR0(i,n) wd[x[i]=r[i]]++;

    FOR1(i,m-1) wd[i]+=wd[i-1];

    FORL0(i,n-1) sa[--wd[x[i]]]=i;

    for(j=1,p=1;p<n;j<<=1,m=p)

    {

        p=0;

        FOR(i,n-j,n-1) y[p++]=i;

        FOR0(i,n) if(sa[i]>=j) y[p++]=sa[i]-j;

        FOR0(i,m) wd[i]=0;

        FOR0(i,n) wd[x[i]]++;

        FOR1(i,m-1) wd[i]+=wd[i-1];

        FORL0(i,n-1) sa[--wd[x[y[i]]]]=y[i];

        t=x;x=y;y=t;p=1;x[sa[0]]=0;

        FOR1(i,n-1) x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;

    }

}







void calHeight(int *r,int *sa,int n)

{

    int i,j,k=0;

    FOR1(i,n) rank[sa[i]]=i;

    FOR0(i,n)

    {

        if(k) k--;

        j=sa[rank[i]-1];

        while(i+k<n&&j+k<n&&r[i+k]==r[j+k]) k++;

        h[rank[i]]=k;

    }

}





int n,m,b[N];

char s[N];









int OK(int x)

{

    int i,j,k,t,L,R,hash[105];

    FOR1(i,m)

    {

        L=i;

        while(L<=m&&h[L]<x) L++;

        if(L>m) break;

        R=L;

        while(R<=m&&h[R]>=x) R++;

        clr(hash,0);

        FOR(j,L-1,R-1) if(b[sa[j]]!=-1)

        {

            hash[b[sa[j]]]=1;

        }

        FOR1(j,n) if(!hash[j]) break;

        if(j>n) return 1;

        i=R;

    }

    return 0;

}





void deal()

{

    int low=0,high=100,mid;

    while(low<=high)

    {

        mid=(low+high)>>1;

        if(OK(mid)) low=mid+1;

        else high=mid-1;

    }

    if(low<=100&&OK(low)) PR(low);

    else if(high>=0&&OK(high)) PR(high);

}



int main()

{

    int C;

    RD(C);

    while(C--)

    {

        RD(n);

        int i,j,L=0,len,Max=130;

        FOR1(i,n)

        {

            RD(s); len=strlen(s);

            FOR0(j,len) r[L++]=s[j],b[L-1]=i;

            r[L++]=Max++;

            b[L-1]=-1;

            FORL0(j,len-1) r[L++]=s[j],b[L-1]=i;

            r[L++]=Max++;

            b[L-1]=-1;

        }

        r[L]=0;

        da(r,sa,L+1,Max);

        calHeight(r,sa,L);

        m=L;

        deal();

    }

    return 0;

}