HDU 5384——Danganronpa——————【AC自动机】

Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 582    Accepted Submission(s): 323

Problem Description

Danganronpa is a video game franchise created and developed by Spike Chunsoft, the series' name is compounded from the Japanese words for "bullet" (dangan) and "refutation" (ronpa).

Now, Stilwell is playing this game. There are n verbal evidences, and Stilwell has m "bullets". Stilwell will use these bullets to shoot every verbal evidence.

Verbal evidences will be described as some strings Ai, and bullets are some strings Bj. The damage to verbal evidence Ai from the bullet Bj is f(Ai,Bj).

f(A,B)=∑i=1|A|−|B|+1[ A[i...i+|B|−1]=B ]

In other words, f(A,B) is equal to the times that string B appears as a substring in string A.
For example: f(ababa,ab)=2, f(ccccc,cc)=4

Stilwell wants to calculate the total damage of each verbal evidence Ai after shooting all m bullets Bj, in other words is ∑mj=1f(Ai,Bj).

 

 

Input

The first line of the input contains a single number T, the number of test cases.
For each test case, the first line contains two integers n, m.
Next n lines, each line contains a string Ai, describing a verbal evidence.
Next m lines, each line contains a string Bj, describing a bullet.

T≤10
For each test case, n,m≤105, 1≤|Ai|,|Bj|≤104, ∑|Ai|≤105, ∑|Bj|≤105
For all test case, ∑|Ai|≤6∗105, ∑|Bj|≤6∗105, Ai and Bj consist of only lowercase English letters

 

 

Output

For each test case, output n lines, each line contains a integer describing the total damage of Ai from all m bullets, ∑mj=1f(Ai,Bj).

 

 

Sample Input

1

5 6

orz

sto

kirigiri

danganronpa

ooooo

o

kyouko

dangan

ronpa

ooooo

ooooo

 

 

Sample Output

1

1

0

3

7

 

 

题目大意：t组测试数据，下面有n个文本串，m个模式串，问每个文本串能匹配多少个模式串。

 

解题思路：AC自动机，记录模式串单词出现的次数cnt，然后直接套模板就可以了。

 

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+200;
char Text[maxn][maxn/10],Pattern[maxn];
struct ACnode{
    ACnode *fail;
    ACnode *next[26];
    int cnt;
    ACnode (){
        fail=NULL;
        for(int i=0;i<26;i++)
            next[i]=NULL;
    }
}*Q[maxn*10];
ACnode *newacnode(){
    ACnode *tmp;
    tmp=new ACnode;
    tmp->cnt=0;
}
void Insert(ACnode *rt,char *s){
    int len=strlen(s);
    int idx;
    for(int i=0;i<len;i++){
        idx=s[i]-'a';
        if(rt->next[idx]==NULL){
            rt->next[idx]=newacnode();
        }
        rt=rt->next[idx];
    }
    rt->cnt++;
}
void BuildAC(ACnode *root){
    int head,tail;
    head=tail=0;
    for(int i=0;i<26;i++){
        if(root->next[i]!=NULL){
            root->next[i]->fail=root;
            Q[tail++]=root->next[i];
        }
    }
    ACnode *p,*tmp;
    while(head!=tail){
        p=Q[head++];
        tmp=NULL;
        for(int i=0;i<26;i++){
            if(p->next[i]!=NULL){
                tmp=p->fail;
                while(tmp!=NULL){
                    if(tmp->next[i]!=NULL){
                        p->next[i]->fail=tmp->next[i];
                        break;
                    }
                    tmp=tmp->fail;
                }
                if(tmp==NULL){
                    p->next[i]->fail=root;
                }
                Q[tail++]=p->next[i];
            }
        }
    }
}
int Query(ACnode *root,char *str){
    ACnode *p=root,*tmp=NULL;
    int idx,len,ret;
    ret=0;
    len=strlen(str);
    for(int i=0;i<len;i++){
        idx=str[i]-'a';
        while( p->next[idx]==NULL && p!=root  )
            p=p->fail;
        p=p->next[idx];
        if(p == NULL)
            p=root;
        tmp=p;
        while(tmp!=root ){
            ret+=tmp->cnt;
            tmp=tmp->fail;
        }
    }
    return ret;
}

int main(){
    int n,m,t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        ACnode *root;
        root=newacnode();
        for(int i=0;i<n;i++){
            scanf("%s",Text[i]);
        }
        for(int i=0;i<m;i++){
            scanf("%s",Pattern);
            Insert(root,Pattern);
        }
        BuildAC(root);
        for(int i=0;i<n;i++){
            int ans=Query(root,Text[i]);
            printf("%d\n",ans);
        }
    }
    return 0;
}