Description
The archaeologists are going to decipher a very mysterious ``language". Now, they know many language patterns; each pattern can be treated as a string on English letters (only lower case). As a sub string, these patterns may appear more than one times in a large text string (also only lower case English letters).
What matters most is that which patterns are the dominating patterns. Dominating pattern is the pattern whose appearing times is not less than other patterns.
It is your job to find the dominating pattern(s) and their appearing times.
Input
The entire input contains multi cases. The first line of each case is an integer, which is the number of patterns N, 1N150. Each of the following N lines contains one pattern, whose length is in range [1, 70]. The rest of the case is one line contains a large string as the text to lookup, whose length is up to 106.
At the end of the input file, number `0' indicates the end of input file.
Output
For each of the input cases, output the appearing times of the dominating pattern(s). If there are more than one dominating pattern, output them in separate lines; and keep their input order to the output.
Sample Input
2
aba
bab
ababababac
6
beta
alpha
haha
delta
dede
tata
dedeltalphahahahototatalpha
0
Sample Output
4
aba
2
alpha
haha
題意：
有n個小寫字母組成的字符床和一個文本串 你的任務是找出哪些字符串在文本中出現的次數最多 例如 aba 在ababa中出現2次 但是bab只出現了一次
輸入n 之後n個字符串 長度為1-70 n小於等於150 之後一個文本串 長度最長為10的6次方
輸出出現最多的次數 以及出現最多的字符串為什麼 如果存在多個 按輸入順序排列
http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=36265
思路：
很明顯的ac自動機
[cpp]
#include<stdio.h> #include<string.h>
#include<malloc.h> #include<queue>
#include<set> using namespace std;
struct node { int count,id;
struct node *next[26];
struct node *fail;
void init() { int i;
for(i=0;i<26;i++) next[i]=NULL;
count=0;
fail=NULL;
id=-1; } }*root;
void insert(char *str,int id) {
int len,k;
node *p=root;
len=strlen(str);
for(k=0;k<len;k++)
{ int pos=str[k]-'a';
if(p->next[pos]==NULL) { p->next[pos]=new node;
p->next[pos]->init(); p=p->next[pos];
} else p=p->next[pos];
} p->count++;
p->id=id; } void getfail() { int i;
node *p=root,*son,*temp;
queue<struct node *>que;
que.push(p);
while(!que.empty()) { temp=que.front();
que.pop(); for(i=0;i<26;i++)
{ son=temp->next[i];
if(son!=NULL) {
if(temp==root) {son->fail=root;}
else {
p=temp->fail; while(p)
{ if(p->next[i])
{ son->fail=p->next[i];
break; } p=p->fail;
}
if(!p) son->fail=root; } que.push(son);
} } } } int num[200]; char str[1000000+100];
void query() { int len,i,cnt=0;
len=strlen(str); node *p,*temp;
p=root; for(i=0;i<len;i++) {
int pos=str[i]-'a';
while(!p->next[pos]&&p!=root) p=p->fail;
p=p->next[pos];// if(!p) p=root;//
temp=p;
/*不要用*temp=*p 因為*p表示一個node，而*temp也表示一個node 但是由於*temp沒有分配空間 所以是不能進行賦值的 但是可以用temp指針去指向p*/ while(temp!=root) { if(temp->count>=1) {
if(temp->id!=-1) num[temp->id]++; // temp->count=-1;
} temp=temp->fail;
} } //printf("%d\n",cnt);
} char rem[160][100]; int main() {
int cas,n; while(scanf("%d",&n)!=EOF)
{ if(!n) break; root=new node;
root->init(); root->fail=NULL;
int i; getchar();
for(i=0;i<n;i++) { gets(rem[i]);
insert(rem[i],i); } getfail();
memset(num,0,sizeof(num));
gets(str); query(); int maxnum=-1;
for(i=0;i<n;i++) {
// printf("num[%d]=%d\n",i,num[i]);
if(num[i]>maxnum) maxnum=num[i];
} printf("%d\n",maxnum);
for(i=0;i<n;i++) if(maxnum==num[i]) printf("%s\n",rem[i]);
} return 0; } #include<stdio.h>
#include<string.h>
#include<malloc.h>
#include<queue>
#include<set>
using namespace std;
struct node
{
int count,id;
struct node *next[26];
struct node *fail;
void init()
{
int i;
for(i=0;i<26;i++)
next[i]=NULL;
count=0;
fail=NULL;
id=-1;
}
}*root;
void insert(char *str,int id)
{
int len,k;
node *p=root;
len=strlen(str);
for(k=0;k<len;k++)
{
int pos=str[k]-'a';
if(p->next[pos]==NULL)
{
p->next[pos]=new node;
p->next[pos]->init();
p=p->next[pos];
}
else
p=p->next[pos];
}
p->count++;
p->id=id;
}
void getfail()
{
int i;
node *p=root,*son,*temp;
queue<struct node *>que;
que.push(p);
while(!que.empty())
{
temp=que.front();
que.pop();
for(i=0;i<26;i++)
{
son=temp->next[i];
if(son!=NULL)
{
if(temp==root) {son->fail=root;}
else
{
p=temp->fail;
while(p)
{
if(p->next[i])
{
son->fail=p->next[i];
break;
}
p=p->fail;
}
if(!p) son->fail=root;
}
que.push(son);
}
}
}
}
int num[200];
char str[1000000+100];
void query()
{
int len,i,cnt=0;
len=strlen(str);
node *p,*temp;
p=root;
for(i=0;i<len;i++)
{
int pos=str[i]-'a';
while(!p->next[pos]&&p!=root) p=p->fail;
p=p->next[pos];//
if(!p) p=root;//
temp=p;
/*不要用*temp=*p 因為*p表示一個node，而*temp也表示一個node 但是由於*temp沒有分配空間 所以是不能進行賦值的 但是可以用temp指針去指向p*/
while(temp!=root)
{
if(temp->count>=1)
{
if(temp->id!=-1)
num[temp->id]++;
// temp->count=-1;
}
temp=temp->fail;
}
}
//printf("%d\n",cnt);
}
char rem[160][100];
int main()
{
int cas,n;
while(scanf("%d",&n)!=EOF)
{
if(!n) break;
root=new node;
root->init();
root->fail=NULL;
int i;
getchar();
for(i=0;i<n;i++)
{
gets(rem[i]);
insert(rem[i],i);
}
getfail();
memset(num,0,sizeof(num));
gets(str);
query();
int maxnum=-1;
for(i=0;i<n;i++)
{
// printf("num[%d]=%d\n",i,num[i]);
if(num[i]>maxnum) maxnum=num[i];
}
printf("%d\n",maxnum);
for(i=0;i<n;i++)
if(maxnum==num[i]) printf("%s\n",rem[i]);
}
return 0;
}
