HDU 4416 Good Article Good sentence （后缀自动机）

 

                                                                                                                                                  Good Article Good sentence

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?

InputThe 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'.
OutputFor each case, print the case number and the number of substrings that ZengXiao Xian can find.

Sample Input

3
2
abab
ab
ba
1
aaa
bbb
2
aaaa
aa
aaa

Sample Output

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

题意为是第一个串的子串，但不是后面所有串的子串的数量。
分析：先用第一个串构建后缀自动机，然后让之后每个串都在自动机上跑一遍，记录在每个节点的最大匹配长度
对于每个节点，它能提供的子串数是p->step-p->link->step。那么现在求满足条件的子串的话，
可以设maxx为该节点被匹配的最大长度,如果maxx>0，该节点的贡献就是p->step-p->maxx(如果maxx>0那么maxx，必然大于p->link->step)
如果maxx=0,那么该点贡献为p->step-p->link->step.最后累加即可
代码如下：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>

using namespace std;
const int N=100010;
typedef long long ll;
struct State
{
    State *link,*go[26];
    int step;
    int maxx;
    void clear()
    {
        maxx=0;
        link=0;
        step=0;
        memset(go,0,sizeof(go));
    }
}*root,*last;
int cnt[N];
int lenA;
State statePool[N*2],*b[N*2],*cur;

void init()
{
    cur=statePool;
    root=last=cur++;
    root->clear();
}

void Insert(int w)
{
    State *p=last;
    State *np=cur++;
    np->clear();
    np->step=p->step+1;
    while(p&&!p->go[w])
        p->go[w]=np,p=p->link;
    if(p==0)
        np->link=root;
    else
    {
        State *q=p->go[w];
        if(p->step+1==q->step)
            np->link=q;
        else
        {
            State *nq=cur++;
            nq->clear();
            memcpy(nq->go,q->go,sizeof(q->go));
            nq->step=p->step+1;
            nq->link=q->link;
            q->link=nq;
            np->link=nq;
            while(p&&p->go[w]==q)
                p->go[w]=nq, p=p->link;
        }
    }
    last=np;
}

void tsort()
{
    memset(cnt,0,sizeof(cnt));
    State *p;
    for(p=statePool;p!=cur;p++)
      cnt[p->step]++;
    for(int i=1;i<=lenA;i++)
      cnt[i]+=cnt[i-1];
    for(p=statePool;p!=cur;p++)
      b[--cnt[p->step]]=p;
}
char A[N],B[N];
int main()
{
    int t,n,len1,Case=0;
    scanf("%d",&t);
    while(t--)
    {
        Case++;
      State *p;
      scanf("%d",&n);
      scanf("%s",A);
      lenA=strlen(A);
      init();
      for(int i=0;i<lenA;i++)
        Insert(A[i]-'a');
      tsort();
      while(n--)
      {
         scanf("%s",B);
         int lenB=strlen(B);
         len1=0;
         p=root;
         for(int i=0;i<lenB;i++)
         {
           int x=B[i]-'a';
           if(p->go[x])
           {
               len1++;
               p=p->go[x];
           }
           else
           {
               while(p&&!p->go[x])p=p->link;
               if(p==NULL)
               {
                  p=root;
                  len1=0;
               }
               else
               {
                  len1=p->step+1;
                  p=p->go[x];
               }
           }
             p->maxx=max(p->maxx,len1);
         }
      }
      int L=cur-statePool;
      ll ans=0;
      for(int i=L-1;i>0;i--)
      {
          p=b[i];
          if(p->maxx>0)
          {
             ans+=p->step-p->maxx;
             p->link->maxx=p->link->step;
          }
          else ans+=p->step-p->link->step;
      }
      printf("Case %d: ",Case);
      printf("%lld
",ans);
    }
    return 0;
}