后缀自动机（不同子串的个数）hdu4416

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Good Article Good sentence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2503    Accepted Submission(s): 710

Problem Description

In middle school, teachers used to encourage us to pick up pretty sentences so that we could apply those sentences in our own articles. One of my classmates ZengXiao Xian, wanted to get sentences which are different from that of others, because he thought the distinct pretty sentences might benefit him a lot to get a high score in his article.
Assume that all of the sentences came from some articles. ZengXiao Xian intended to pick from Article A. The number of his classmates is n. The i-th classmate picked from Article Bi. Now ZengXiao Xian wants to know how many different sentences she could pick from Article A which don't belong to either of her classmates?Article. To simplify the problem, ZengXiao Xian wants to know how many different strings, which is the substring of string A, but is not substring of either of string Bi. Of course, you will help him, won't you?

 

Input

The first line contains an integer T, the number of test data. 
For each test data
The first line contains an integer meaning the number of classmates.
The second line is the string A;The next n lines,the ith line input string Bi.
The length of the string A does not exceed 100,000 characters , The sum of total length of all strings Bi does not exceed 100,000, and assume all string consist only lowercase characters 'a' to 'z'.

 

Output

For each case, print the case number and the number of substrings that ZengXiao Xian can find.

 

Sample Input

32abababba1aaabbb2aaaaaaaaa

 

Sample Output

Case 1: 3Case 2: 3Case 3: 1

对A串建立后缀自动机，然后用其他的串去匹配，每个节点维护一个cnt，表示最大匹配长度，最后统计一下，不同的子串数量就可以了

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;typedef long long LL;const int maxn=100010;const int SIGMA_SIZE=26;struct SAM_Node{    SAM_Node *par,*next[SIGMA_SIZE];    int len,id,pos;    int cnt;    SAM_Node() {}    SAM_Node(int _len)    {        par=0;        len=_len;        cnt=0;        memset(next,0,sizeof(next));    }};SAM_Node node[maxn*2],*root,*last;int SAM_size;SAM_Node *newSAM_Node(int len){    node[SAM_size]=SAM_Node(len);    node[SAM_size].id=SAM_size;    return &node[SAM_size++];}SAM_Node *newSAM_Node(SAM_Node *p){    node[SAM_size]=*p;    node[SAM_size].id=SAM_size;    return &node[SAM_size++];}void SAM_add(int x,int len){    SAM_Node *p=last,*np=newSAM_Node(p->len+1);    np->pos=len;    last=np;    while(p&&!p->next[x])    {        p->next[x]=np;        p=p->par;    }    if(!p)np->par=root;    else    {        SAM_Node *q=p->next[x];        if(q->len==p->len+1)            np->par=q;        else        {            SAM_Node *nq=newSAM_Node(q);            nq->len=p->len+1;            q->par=nq;            np->par=nq;            while(p&&p->next[x]==q)            {                p->next[x]=nq;                p=p->par;            }        }    }}void SAM_init(){    SAM_size=0;    root=last=newSAM_Node(0);    node[0].pos=0;}void SAM_build(char *s){    SAM_init();    int len=strlen(s);    for(int i=0;i<len;i++)        SAM_add(s[i]-'a',i+1);}char s[maxn];int N;int cnt[maxn*2];SAM_Node *sa[maxn*2];int main(){    int T,cas=1;    scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        scanf("%s",s);        SAM_build(s);        int len=strlen(s);        for(int i=0;i<=len;i++)cnt[i]=0;        for(int i=0;i<SAM_size;i++)cnt[node[i].len]++;        for(int i=1;i<=len;i++)cnt[i]+=cnt[i-1];        for(int i=SAM_size-1;i>=0;i--)sa[--cnt[node[i].len]]=&node[i];        LL ans=0;        while(N--)        {            scanf("%s",s);            SAM_Node *p=root;            int num=0;            len=strlen(s);            for(int i=0;i<len;i++)            {                int c=s[i]-'a';                if(p->next[c])num++,p=p->next[c];                else                {                    while(p&&!p->next[c])p=p->par;                    if(p){num=p->len+1;p=p->next[c];}                    else{num=0,p=root;}                }                p->cnt=max(p->cnt,num);            }        }        for(int i=SAM_size-1;i>=0;i--)        {            SAM_Node *p=sa[i];            if(p->cnt>0)            {                if(p->par)p->par->cnt=max(p->par->cnt,p->cnt);                if(p->cnt<p->len)ans+=p->len-p->cnt;            }            else ans+=p->len-(p->par?p->par->len:0);        }        printf("Case %d: %I64d\n",cas++,ans);    }    return 0;}