hdu 6096 ac自动机 和 处理字符串分段输入

Bob has a dictionary with N words in it. 

Now there is a list of words in which the middle part of the word has continuous letters disappeared. The middle part does not include the first and last character. 

We only know the prefix and suffix of each word, and the number of characters missing is uncertain, it could be 0. But the prefix and suffix of each word can not overlap. 

For each word in the list, Bob wants to determine which word is in the dictionary by prefix and suffix. 

There are probably many answers. You just have to figure out how many words may be the answer.

InputThe first line of the input gives the number of test cases T; T test cases follow. 

Each test case contains two integer N and Q, The number of words in the dictionary, and the number of words in the list. 

Next N line, each line has a string Wi, represents the ith word in the dictionary (0<|Wi|≤1000000<|Wi|≤100000) 

Next Q line, each line has two string Pi , Si, represents the prefix and suffix of the ith word in the list (0<|Pi|,|Si|≤100000,0<|Pi|+|Si|≤1000000<|Pi|,|Si|≤100000,0<|Pi|+|Si|≤100000) 

All of the above characters are lowercase letters. 

The dictionary does not contain the same words. 

Limits 
T≤5T≤5 
0<N,Q≤1000000<N,Q≤100000 
∑Si+Pi≤500000∑Si+Pi≤500000 
∑Wi≤500000∑Wi≤500000 

OutputFor each test case, output Q lines, an integer per line, represents the answer to each word in the list. 

Sample Input

1
4 4
aba
cde
acdefa
cdef
a a
cd ef
ac a
ce f

Sample Output

2
1
1

0

ac自动机 最新板子解法

#include <bits/stdc++.h>
using namespace std;
#define maxm 505000*3
const int M = 505000*3;
char str[maxm];
int ans[M];
const int SIGMA_SIZE=27;

struct AC
{
int ch[maxm][27];
int val[maxm];
int height[maxm];
int fail[maxm],last[maxm];
int sz;
void clear(){memset(ch[0],0,sizeof(ch[0]));sz=1;}
int idx(char x){return x-'a';}
int insert(char *s)
{
int u=0;
int n=strlen(s);
for(int i=0;i<n;i++)
{
int c=idx(s[i]);
if(!ch[u][c])
{
memset(ch[sz],0,sizeof(ch[sz]));
height[sz]=height[u]+1;
val[sz]=0;
ch[u][c]=sz++;
}
u=ch[u][c];
}
val[u]++;
return u;
}
void getfail()
{
queue<int> q;
fail[0]=0;
int u=0;
for(int i=0;i<SIGMA_SIZE;i++)
{
u=ch[0][i];
if(u){q.push(u);fail[u]=0;last[u]=0;}
}
while(!q.empty())
{
int r=q.front();q.pop();
for(int i=0;i<SIGMA_SIZE;i++)
{
u=ch[r][i];
if(!u){ch[r][i]=ch[fail[r]][i];continue;}
q.push(u);
int v=fail[r];
while(v&&!ch[v][i])v=fail[v];
fail[u]=ch[v][i];
last[u]=val[fail[u]]?fail[u]:last[fail[u]];
}
}
}
int find(char *s,int le)
{
int u=0,cnt=0;
for(int i=0;s[i];i++)
{
int c=idx(s[i]);
u=ch[u][c];
int temp=0;//必须赋初值为0，表示如果下面两个判断都不成立的时候while可以正常执行
if(val[u])
temp=u;
else if(last[u])
temp=last[u];
while(temp)
{
if(height[temp]<=le)
ans[temp]++;
temp=last[temp];
}
}
}
}tree;

int d[M],st[M];

const int N = 100100;
int len
;
char *s
;
int pos
;
char ss[M],t1[M],t2[M];

int main()
{
int T,n,i;
scanf("%d",&T);
while(T--){
int n,m;
tree.clear();
memset(ans,0,sizeof(ans));
scanf("%d%d",&n,&m);
int del=0;
int j=0;
for(int i=1;i<=n;i++)
{
s[i]=ss+j;
scanf(" %s",s[i]);
len[i]=strlen(s[i])+1;
j+=len[i];
strcpy(ss+j,s[i]);
ss[j-1]='z'+1;
j+=len[i];
}

for(int i=1;i<=m;i++)
{
scanf(" %s %s",t1+1,t2);
t1[0]='z'+1;
strcat(t2,t1);
pos[i]=tree.insert(t2);
}
tree.getfail();

for(int i=1;i<=n;i++)
{
tree.find(s[i],len[i]);
}

for(int i=1;i<=m;i++) printf("%d\n",ans[pos[i]] );
}
return 0;
}