Minimum representation of poj 1509 strings (SAM suffix automatic machine)

Http://poj.org/problem? Id = 1509

It is the position of the starting character in the minimum notation of a string. If there are multiple characters, the minimum position is obtained.

 

Idea: This question is self-idle with the suffix of the automatic machine. It is a little tricky. This question has other more streamlined algorithms, and the time complexity is O (n ), but here Sam can be used to have a clearer understanding of Sam, simply to practice Sam.

Copy the original character s string and add it to the original string and convert it to SS. Then, we need the smallest lexicographic string with the length of SS | S |, so we can construct the SS Sam, starting from root, each time the smallest number is transferred, | S | after the step arrives, it is the requested substring (set to P ), because we require the starting position and the smallest one, we save a variable L in each State to indicate the leftmost position of this State in SS, the answer is P-> L-| S | + 1. below is the code.

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#define maxn 40010using namespace std;char str[maxn>>1];struct sam{    sam *go[26],*par;    int val;    int l,po;}*root,*tail,que[maxn],*top[maxn];int tot;void add(int c,int l,int po){    sam *p=tail,*np=&que[tot++];    np->po=po;    np->val=l;    while(p&&p->go[c]==NULL)    p->go[c]=np,p=p->par;    if(p==NULL) np->par=root;    else    {        sam *q=p->go[c];        if(p->val+1==q->val) np->par=q;        else        {            sam *nq=&que[tot++];            *nq=*q;            nq->val=p->val+1;            np->par=q->par=nq;            while(p&&p->go[c]==q) p->go[c]=nq,p=p->par;        }    }    tail=np;}int c[maxn],len;void init(){    memset(que,0,sizeof(que));    tot=0;    len=1;    root=tail=&que[tot++];}void solve(int n){    memset(c,0,sizeof(c));    int i;    for(i=0;i<tot;i++)    c[que[i].val]++;    for(i=1;i<len;i++)    c[i]+=c[i-1];    for(i=0;i<tot;i++)    top[--c[que[i].val]]=&que[i];    for(sam *p=root;;p=p->go[str[p->val+1]-'a'])    {        p->l=p->po;        //printf("%c",str[p->val+1]);        if (p->val==len-1)break;    }    sam *p;    for(i=tot-1;i>=0;i--)    {        p=top[i];        if(p->par)        {            sam *q=p->par;            if(q->l==0||q->l>p->l)            q->l=p->l;        }    }    p=root;    int tmp=n;    while(tmp--)    {        int i;        for(i=0;i<26;i++)        {            if(p->go[i])            {                p=p->go[i];                break;            }        }    }    int ans=p->l-n+1;    printf("%d\n",ans);}int main(){    //freopen("dd.txt","r",stdin);    int ncase;    scanf("%d",&ncase);    while(ncase--)    {        scanf("%s",str+1);        init();        int l=strlen(str+1),i;        for(i=1;i<=l;i++)        {            str[i+l]=str[i];        }        for(i=1;i<=2*l;i++)        add(str[i]-'a',len++,i);        solve(l);    }    return 0;}

Here is another code specifically used to calculate the minimum string representation. Simple and effective...

#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAXN 100000int min(int a,int b){    return a<b?a:b;}int min_Str(char *str){    int i,j,k,len=strlen(str);    i=0,j=1;    memcpy(str+len,str,len);    while(j<len&&i<len)    {        k=0;        while(str[i+k]==str[j+k])        k++;        if(k>=len)        break;        if(str[i+k]>str[j+k])        i=i+k+1;        else        j=j+k+1;        if(i==j)        j++;    }    return min(i,j);}char str[MAXN];int main(){   //freopen("dd.txt","r",stdin);    int ncase;    scanf("%d",&ncase);    while(ncase--)    {   memset(str,'\0',sizeof(str));        scanf("%s",str);        printf("%d\n",1+min_Str(str));    }    return 0;}