hdu5384 Danganronpa AC自动机 多校联合第八场

Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 13    Accepted Submission(s): 1

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
   
   
   
   

  先输入N个串，再输入M个串，对前N个串中的每个串输出后面M个串在它上面匹配了多少次。

  把后M个串建立AC自动机，val[u]表示以u这个结点为结尾时匹配了几个串，别忘了val[u]+=val[f[u]]。然后对前N个串分别find，复杂度是所有字符串的长度和。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#define INF 0x3f3f3f3f
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

typedef long long LL;

const int SIGMA_SIZE=26;
const int MAXN=100010;
const int MAXNODE=600010;
const LL MOD= 258280327;

int T,N,M;
LL ans;
string str1[MAXN],str2[MAXN];

struct AC{
    int ch[MAXNODE][SIGMA_SIZE],f[MAXNODE],sz;
    LL val[MAXNODE];
    void init(){
        memset(ch[0],0,sizeof(ch[0]));
        val[0]=0;
        sz=1;
    }
    int idx(char c){
        return c-'a';
    }
    void insert(string s){
        int u=0,len=s.size();
        for(int i=0;i<len;i++){
            int c=idx(s[i]);
            if(!ch[u][c]){
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz]=0;
                ch[u][c]=sz++;
            }
            u=ch[u][c];
        }
        val[u]++;
    }
    void get_fail(){
        queue<int> q;
        f[0]=0;
        for(int c=0;c<SIGMA_SIZE;c++){
            int u=ch[0][c];
            if(u){
                f[u]=0;
                q.push(u);
            }
        }
        while(!q.empty()){
            int r=q.front();
            q.pop();
            for(int c=0;c<SIGMA_SIZE;c++){
                int u=ch[r][c];
                if(!u){
                    ch[r][c]=ch[f[r]][c];
                    continue;
                }
                q.push(u);
                f[u]=ch[f[r]][c];
                val[u]+=val[f[u]];
            }
        }
    }

    void find(string T){
        int j=0;
        int len=T.size();
        for(int i=0;i<len;i++){
            int c=idx(T[i]);
            j=ch[j][c];
            ans+=val[j];
        }
    }
}ac;

int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&N,&M);
        for(int i=1;i<=N;i++) cin>>str1[i];
        for(int i=1;i<=M;i++) cin>>str2[i];
        ac.init();
        for(int i=1;i<=M;i++) ac.insert(str2[i]);
        ac.get_fail();
        for(int i=1;i<=N;i++){
            ans=0;
            ac.find(str1[i]);
            printf("%I64d\n",ans);
        }
    }
    return 0;
}