二次联通门 : luogu P2353 背单词
一眼看过去， 卧槽，AC自动机板子题
写完后T成SB
卧槽10^6 做个篮子啊
重构思路。。。
恩。。Hash + 莫队。。。
恶心啊。。
找xxy dalao， AC自动机 + 前缀和
码完WA成SB
去群里找dalao
大佬告诉了我前缀和的正确使用姿势。。。
然后就依然WA成SB
做个毛线
贴一个AC自动机的30代码
#include <cstdio>
#include <cstring>
#include <queue>
#define Max 3000090
void read (int &now)
{
now = 0;
register char word = getchar ();
while (word < ‘0‘ || word > ‘9‘)
word = getchar ();
while (word >= ‘0‘ && word <= ‘9‘)
{
now = now * 10 + word - ‘0‘;
word = getchar ();
}
}
char __txt[Max];
struct T_D
{
T_D *child[26];
T_D *Fail;
int Count;
int number;
T_D ()
{
for (int i = 0; i < 26; i ++)
this->child[i] = NULL;
Count = 0;
Fail = NULL;
number = 0;
}
};
int List[Max << 1];
int List_Cur;
bool visit[Max];
class AC_Type
{
private :
T_D *Root;
int Trie_Count;
public :
void Insert (char *key)
{
T_D *now = Root;
int Len = strlen (key);
int Id;
for (int i = 0; i < Len; i ++)
{
Id = key[i] - ‘a‘;
if (now->child[Id] == NULL)
{
now->child[Id] = new T_D;
now->child[Id]->number = ++ Trie_Count;
}
now = now->child[Id];
}
now->Count ++;
}
AC_Type ()
{
Trie_Count = 0;
Root = new T_D ;
Root->number = ++ Trie_Count;
}
void Build_AC ()
{
std :: queue <T_D *> Queue;
Queue.push (Root);
T_D *now, *pos;
while (!Queue.empty ())
{
now = Queue.front ();
Queue.pop ();
pos = NULL;
for (int i = 0; i < 26; i ++)
{
if (now->child[i] == NULL)
continue;
if (now == Root)
now->child[i]->Fail = Root;
else
{
for (pos = now->Fail; pos; pos = pos->Fail)
if (pos->child[i])
{
now->child[i]->Fail = pos->child[i];
break;
}
if (pos == NULL)
now->child[i]->Fail = Root;
}
Queue.push (now->child[i]);
}
}
}
int Query (int x, int y)
{
T_D *now, *pos;
int Id ;
now = Root;
int res = 0;
for (int i = x; i <= y; i ++)
{
Id = __txt[i] - ‘a‘;
for (; now != Root && now->child[Id] == NULL; now = now->Fail);
now = now->child[Id];
if (now == NULL)
now = Root;
for (pos = now; pos != Root && !visit[pos->number]; pos = pos->Fail)
{
res += pos->Count;
visit[pos->number] = true;
List[++ List_Cur] = pos->number;
}
for (int j = 1; j <= List_Cur; j ++)
visit[List[j]] = false;
List_Cur = 0;
}
return res;
}
};
AC_Type Make;
int N, Q;
char line[Max];
int main (int argc, char *argv[])
{
read (N);
read (Q);
scanf ("%s", __txt);
for (int i = 1; i <= N; i ++)
{
scanf ("%s", line);
Make.Insert (line);
}
Make.Build_AC ();
for (int x, y; Q --; )
{
read (x);
read (y);
printf ("%d\n", Make.Query (-- x, -- y));
}
return 0;
}
再贴个AC自动机思路正确但由于细节问题WA成dog的代码
#include <cstdio>
#include <cstring>
#include <queue>
#define Max 1000090
#define DEBUG for (int i = 1; i <= strlen (__txt); i ++)\
printf ("%d ", __sum[i]); putchar (‘\n‘);
void read (int &now)
{
now = 0;
register char word = getchar ();
while (word < ‘0‘ || word > ‘9‘)
word = getchar ();
while (word >= ‘0‘ && word <= ‘9‘)
{
now = now * 10 + word - ‘0‘;
word = getchar ();
}
}
char __txt[Max];
struct T_D
{
T_D *child[26];
T_D *Fail;
int Count;
int number;
T_D ()
{
for (int i = 0; i < 26; i ++)
this->child[i] = NULL;
Count = 0;
Fail = NULL;
number = 0;
}
};
int __sum[Max];
class AC_Type
{
private :
T_D *Root;
int Trie_Count;
public :
void Insert (char *key)
{
T_D *now = Root;
int Len = strlen (key);
int Id;
for (register int i = 0; i < Len; i ++)
{
Id = key[i] - ‘a‘;
if (now->child[Id] == NULL)
{
now->child[Id] = new T_D;
now->child[Id]->number = ++ Trie_Count;
}
now = now->child[Id];
}
now->Count ++;
}
AC_Type ()
{
Trie_Count = 0;
Root = new T_D ;
Root->number = ++ Trie_Count;
}
void Build_AC ()
{
std :: queue <T_D *> Queue;
Queue.push (Root);
T_D *now, *pos;
while (!Queue.empty ())
{
now = Queue.front ();
Queue.pop ();
pos = NULL;
for (register int i = 0; i < 26; i ++)
{
if (now->child[i] == NULL)
continue;
if (now == Root)
now->child[i]->Fail = Root;
else
{
for (pos = now->Fail; pos; pos = pos->Fail)
if (pos->child[i])
{
now->child[i]->Fail = pos->child[i];
break;
}
if (pos == NULL)
now->child[i]->Fail = Root;
}
Queue.push (now->child[i]);
}
}
}
int Query ()
{
T_D *now, *pos;
int Id ;
now = Root;
int res = 0;
int Len = strlen (__txt);
for (register int i = 0; i < Len; i ++)
{
Id = __txt[i] - ‘a‘;
for ( ; now != Root && now->child[Id] == NULL; now = now->Fail);
now = now->child[Id];
if (now == NULL)
now = Root;
for (pos = now; pos != Root && pos->Count >= 0; pos = pos->Fail)
{
__sum[i + 1] += pos->Count;
pos->Count = -1;
}
__sum[i + 1] += __sum[i];
}
return res;
}
};
AC_Type Make;
int N, Q;
char line[Max];
int length[Max];
int main (int argc, char *argv[])
{
read (N);
read (Q);
scanf ("%s", __txt);
for (int i = 1; i <= N; i ++)
{
scanf ("%s", line);
Make.Insert (line);
length[i] = strlen (line);
}
Make.Build_AC ();
Make.Query ();
register int Answer, now;
for (int x, y; Q --; )
{
Answer = 0;
read (x);
read (y);
for (int i = 1; i <= N; i ++)
Answer += (__sum[y - length[i]] - __sum[x - 1]);
printf ("%d\n", Answer);
}
return 0;
}
最后再贴个正解。。。。是我想麻烦了。。kmp或者hash都可以。。
时间： 2024-12-16 02:40:23