hdu5384 AC自動機模板題，統計模式串在給定串中出現的個數

life4711發表於2015-08-14
原文網址 : https://blog.csdn.net/lvshubao1314/article/details/47667927
模式

http://acm.hdu.edu.cn/showproblem.php?pid=5384

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








/**
hdu5384 AC自動機模板題，統計模式串在給定串中出現的個數
*/
#include<string.h>
#include<algorithm>
#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
char str[100010][10010];
int num[100010],n,m;
struct Trie
{
    int next[10010*50][28],fail[10010*50],end[10010*50];
    int root,L;
    int newnode()
    {
        for(int i=0; i<26; i++)
        {
            next[L][i]=-1;
        }
        end[L++]=-1;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    void insert(char *s)
    {
        int len=strlen(s);
        int now=root;
        for(int i=0; i<len; i++)
        {
            if(next[now][s[i]-'a']==-1)
            {
                next[now][s[i]-'a']=newnode();
            }
            now=next[now][s[i]-'a'];
        }
        if(end[now]==-1)///標記模式串出現的次數
        {
            end[now]=1;
        }
        else
        {
            end[now]++;
        }
    }
    void build()
    {
        queue<int>Q;
        fail[root]=root;
        for(int i=0; i<26; i++)
        {
            if(next[root][i]==-1)
            {
                next[root][i]=root;
            }
            else
            {
                fail[next[root][i]]=root;
                Q.push(next[root][i]);
            }
        }
        while(!Q.empty())
        {
           // printf("**\n");
            int now=Q.front();
            Q.pop();
            for(int i=0; i<26; i++)
            {
                if(next[now][i]==-1)
                    next[now][i]=next[fail[now]][i];
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    void query(char* s)
    {
        memset(num,0,sizeof(num));
        int len=strlen(s);
        int now=root;
        for(int i=0; i<len; i++)
        {
            now=next[now][s[i]-'a'];
            int temp=now;
            while(temp!=root)
            {
                if(end[temp]!=-1)///統計所有模式串出現的次數，num陣列在0~m之間定能取到所有end[temp]必不大於m
                {
                    num[end[temp]]+=end[temp];
                }
                temp=fail[temp];
            }
        }
        int ans=0;
        for(int i=0; i<=m; i++)
        {
            if(num[i]>0)
                ans+=num[i];
        }
        printf("%d\n",ans);
    }
} ac;
char s[10005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++)
        {
            scanf("%s",str[i]);
        }
        ac.init();
        for(int i=0; i<m; i++)
        {
            scanf("%s",s);
            ac.insert(s);
        }
        ac.build();
        //printf("**\n");
        for(int i=0; i<n; i++)
        {
            ac.query(str[i]);
        }
    }
    return 0;
}



            




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