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题意:
给一个文本串和多个模式串,求文本串中一共出现多少次模式串
分析:
ac自动机模板,关键是失配函数
#include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include <string> #include <cctype> #include <complex> #include <cassert> #include <utility> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; typedef pair<int,int> PII; typedef long long ll; #define lson l,m,rt<<1 #define pi acos(-1.0) #define rson m+1,r,rt<<11 #define All 1,N,1 #define read freopen("in.txt", "r", stdin) #define N 250010 const ll INFll = 0x3f3f3f3f3f3f3f3fLL; const int INF= 0x7ffffff; const int mod = 1000000007; struct Trie{ int ch[N][26],val[N],f[N],last[N],num; void init(){ num=1; memset(ch,0,sizeof(ch)); memset(val,0,sizeof(val)); memset(f,0,sizeof(f)); memset(last,0,sizeof(last)); } void print(int j){ if(j){ printf("%d: %d\n", j, val[j]); print(last[j]); } } void build(char *s){ int u=0,len=strlen(s); for(int i=0;i<len;++i) { int v=s[i]-‘a‘; if(!ch[u][v]){ memset(ch[num],0,sizeof(ch[num])); ch[u][v]=num++; } u=ch[u][v]; } val[u]++; }
//失配函数 void getfail(){ queue<int>q; for(int i=0;i<26;++i) if(ch[0][i]) q.push(ch[0][i]); while(!q.empty()){ int r=q.front(); q.pop(); for(int i=0;i<26;++i) { int u=ch[r][i]; if(!u)continue; q.push(u); int v=f[r]; while(v&&!ch[v][i])v=f[v]; f[u]=ch[v][i]; last[u]=val[f[u]]?f[u]:last[f[u]]; } } } int find(char *T){ int u=0,len=strlen(T); int total=0; for(int i=0;i<len;++i){ int v=T[i]-‘a‘; while(u&&ch[u][v]==0) u=f[u]; u=ch[u][v]; int tmp=u; while(tmp&&val[tmp]){ total+=val[tmp]; val[tmp]=0; tmp=f[tmp]; } } return total; } }ac; int main() { char s[60],T[1000010]; int t,n; scanf("%d",&t); while(t--){ scanf("%d",&n); ac.init(); while(n--){ scanf("%s",s); ac.build(s); } ac.getfail(); scanf("%s",T); printf("%d\n",ac.find(T)); } return 0; }
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原文地址:http://www.cnblogs.com/zsf123/p/4734569.html