HDU 4416 Good Article Good sentence

参考题解

题目大意：给一个字符串S和一系列字符串T1~Tn，问在S中有多少个不同子串满足它不是T1~Tn中任意一个字符串的子串。

 

思路：我们先构造S的后缀自动机，然后将每一个Ti在S的SAM上做匹配，类似于LCS，在S中的每一个状态记录一个变量deep，表示T1~Tn，在该状态能匹配的最大长度是多少，将每一个Ti匹配完之后，我们将S的SAM做拓扑排序，自底向上更新每个状态的deep，同时计算在该状态上有多少个子串满足题目要求。具体步骤如下：

1：对于当前状态，设为p，设p的par为q，则更新q->deep为q->deep和p->deep中的较大值。

2：若p->deep<p->val,则表示在状态p中，长度为p->deep+1~p->val的子串不是T1~Tn中任意字符串的子串，所以答案加上p->val-p->deep。否则表示状态p中所有字串均不满足要求，跳过即可。

(注意若p->deep==0，表示状态p中所有的子串均满足题目要求，但是答案不是加上p->val-0，而是加上 p->val-p->par->val，这表示状态p中的字符串个数，所以对于p->deep==0要特殊处理)

最后输出答案即可。

Good Article Good sentence

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2152    Accepted Submission(s): 602

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

   
   
   
   

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

 

Sample Output

   
   
   
   

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


   
   
   
   

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define prt(k) cout<<#k"="<<k<<endl;
#define ll long long
const int N=200200;

struct Node
{
    Node *ch[26],*f;
    int len,id,pos;
    Node() {}
    Node(int _len) { len=_len; memset(ch,0,sizeof ch); f=0; }
}sam[N],*root,*last;
int tot; ///SAM_size
Node* newnode(int len)
{
    sam[tot]=Node(len);
    sam[tot].id=tot;
    return &sam[tot++];
}
Node* newnode(Node* p)
{
    sam[tot]=*p;
    sam[tot].id=tot;
    return &sam[tot++];
}
void SAM_init()
{
    tot=0;
    root=last=newnode(0);
    sam[0].pos=0;
}
#define node Node
void add(int x,int len)
{
    Node *p=last,*np=newnode(p->len+1);
    np->pos=len; last=np;
    while(p&&!p->ch[x]) p->ch[x]=np,p=p->f;
    if(!p) { np->f=root;return; }
    node *q=p->ch[x];
    if(q->len==p->len+1) { np->f=q; return; }    ///!!!
    node* nq=newnode(q);
    nq->len=p->len+1;
    q->f=nq; np->f=nq;
    for(;p&&p->ch[x]==q;p=p->f) p->ch[x]=nq;
}
void SAM_build(char s[])
{
    SAM_init();
    for(int i=0;s[i];i++) add(s[i]-'a',i+1);
}
Node* top[N];   ///toposort
char s[N];
int c[N];
int dp[N];
void Max(int& a,int b) { if(a<b) a=b; }
int main()
{
    int re; cin>>re;  int ca=1;
while(re--)
{
    int m; cin>>m;
    scanf("%s",s);
    SAM_build(s);
    memset(c,0,sizeof c);
    memset(dp,0,sizeof dp);
    memset(top,0,sizeof top);
    for(int i=0;i<tot;i++) c[sam[i].len]++;
    for(int i=1;i<=tot;i++) c[i]+=c[i-1];
    for(int i=0;i<tot;i++)
        top[--c[sam[i].len]]=&sam[i];
    while(m--)
    {
        node* p=root;
        scanf("%s",s);
        int len=strlen(s);
        int tmp=0;
        for(int i=0;i<len;i++)
        {
            int x=s[i]-'a';
            if(p->ch[x])
            {
                tmp++;
                p=p->ch[x];
                Max(dp[p->id],tmp);
            }
            else {
                while(p&&!p->ch[x])p=p->f;
                if(p) {
                    tmp=p->len+1;
                    p=p->ch[x];
                    Max(dp[p->id],tmp);
                }
                else { tmp=0; p=root; }
            }
        }
    }
    ll ans=0;
    for(int i=tot-1;i>0;i--)
    {
        Node *p=top[i];
        if(dp[p->id]==0)
        {
            ans+=p->len-p->f->len;
            continue;
        }
        if(p->f) Max(dp[p->f->id],dp[p->id]);
        if(dp[p->id]<p->len) ans+=p->len-dp[p->id];
    }
    printf("Case %d: %I64d\n",ca++,ans);
}
}