AC自动机 白书模板
2016-04-21 19:00
381 查看
模板:
struct ACauto
{
int ch[maxn][26];
int sz;
int f[maxn],last[maxn],val[maxn],cnt[maxn];
void init()
{
sz=1;
memset(ch[0],0,sizeof ch[0]);
memset(cnt,0,sizeof cnt);
}
int idx(char c)
{
return c-'A';
}
void add(char *s,int v)
{
int u=0,len=strlen(s);
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]=v;
}
void getfail()
{
queue<int>q;
f[0]=0;
for(int c=0;c<26;c++)
{
int u=ch[0][c];
if(u)
{
f[u]=0;
q.push(u);
last[u]=0;
}
}
while(!q.empty())
{
int r=q.front();q.pop();
for(int c=0;c<26;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];
last[u]=val[f[u]]?f[u]:last[f[u]];
}
}
}
void print(int j)
{
if(j)
{
cnt[val[j]]++;
print(last[j]);
}
}
void Find(char *T)
{
int n=strlen(T);
int j=0;
for(int i=0;i<n;i++)
{
int c=idx(T[i]);
while(j&&!ch[j][c])j=f[j];
j=ch[j][c];
if(val[j])print(j);
else if(last[j])print(last[j]);
}
}
}ac;
http://acm.hdu.edu.cn/showproblem.php?pid=2222
Keywords Search
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 38991 Accepted Submission(s): 12571
[align=left]Problem Description[/align]In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.
[align=left]Input[/align]First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.
[align=left]Output[/align]Print how many keywords are contained in the description.
[align=left]Sample Input[/align]1
5
she
he
say
shr
her
yasherhs
[align=left]Sample Output[/align]3
[align=left]Author[/align]Wiskey
[align=left]Recommend[/align]lcy
#include<iostream> #include<string.h> #include<stdio.h> #include<string> #include<queue> #define maxn 250010 using namespace std; struct ACauto { int ch[maxn][26]; int sz; int f[maxn],last[maxn],val[maxn][2]; int num; int used[maxn]; void init() { num=0; sz=1; memset(ch[0],0,sizeof(ch[0])); memset(used,0,sizeof used); memset(val,0,sizeof(val)); } int idx(char c) { return c-'a'; } void add(char *s,int v) { int u=0,len=strlen(s); 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]=0; ch[u][c]=sz++; } u=ch[u][c]; } val[u][0]=v;//???????á??±ê?? val[u][1]++; } void print(int j) { if(j&&!used[val[j][0]]) { used[val[j][0]]=1; num+=val[j][1]; print(last[j]); } } void getfail() { queue<int>q; f[0]=0; for(int c=0;c<26;c++) { int u=ch[0][c]; if(u) { f[u]=0; q.push(u); last[u]=0; } } while(!q.empty()) { int r=q.front();q.pop(); for(int c=0;c<26;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]; last[u]=val[f[u]]?f[u]:last[f[u]]; } } } void Find(char*T) { int n=strlen(T); int j=0; for(int i=0;i<n;i++) { int c=idx(T[i]); while(j&&!ch[j][c])j=f[j]; j=ch[j][c]; if(val[j])print(j); else if(last[j])print(last[j]); } } }ac; char s[60],T[1000010]; int main() { int t,n; scanf("%d",&t); while(t--) { ac.init(); scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%s",s); ac.add(s,i); } ac.getfail(); scanf("%s",T); ac.Find(T); printf("%d\n",ac.num); } return 0; }
相关文章推荐
- HDU 1247 Hat’s Words
- HDU 1251 统计难题
- POJ 3630 Phone List
- UVa 11732 - strcmp() Anyone?
- 报名|「OneAPM x DaoCloud」技术公开课:Docker性能监控!
- 基于hexo+github搭建一个独立的博客
- 字典树(trie 树)
- POJ 3984 迷宫问题
- HDU 2612 Find a way
- 问卷2
- C语言qsort的三种使用
- javascript、JavaWeb、URL的区别联系
- intellij idea15 破解
- openstack新建虚机、网络、路由时候对应的ovs网桥的变化
- POJ 1426 Find The Multiple
- HDU 2181 哈密顿绕行世界问题
- IOS中NSUserDefaults的用法
- Codeforces Round #291 (Div. 2)
- Codeforces Round #292 (Div. 2)
- Codeforces Round #293 (Div. 2)