ybtoj【字元串算法】5章3題【字首比對】

字首比對

題目

解析

查詢串考慮字首，故不難想到拿查詢串建，母串跑，跑到的點的 n x t nxt nxt顯然也合法，跳失配标記，就标記出了母串的子串

然後再拿查詢串跑一遍，就好了

code:

#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
inline int f(char s){return (s=='S')?0:((s=='E')?1:((s=='W')?2:3));}
struct AC_easy
{
	int tot=1,q,ch[1000010][4],nxt[1000010],val[1000010],k;
	queue <int> b;
	inline void insert(char *s)
	{
		int len=strlen(s),now=1;
		for(int i=0;i<len;++i)
		{
			if(!ch[now][f(s[i])])ch[now][f(s[i])]=++tot;
			now=ch[now][f(s[i])];
		}
	}
	inline void bfs()
	{
		for(int i=0;i<=3;++i)ch[0][i]=1;
		b.push(1);
		while(b.size())
		{
			k=b.front(),b.pop();
			for(int i=0;i<=3;++i)
			{
				if(!ch[k][i])ch[k][i]=ch[nxt[k]][i];
				else nxt[ch[k][i]]=ch[nxt[k]][i],b.push(ch[k][i]);
			}
		}
	}
	inline void bfs2(char *s)
	{
		int len=strlen(s),now=1,q;
		for(int i=0;i<len;++i){now=q=ch[now][f(s[i])];while(q)val[q]=1,q=nxt[q];}
	}
	inline int ask(char *s)
	{
		int len=strlen(s),now=1,ans=0;
		for(int i=0;i<len;++i)now=ch[now][f(s[i])],ans=val[now]?(i+1):ans;
		return ans;
	}
}AC;
char s[10000010],q[100010][110];
int n;
int main()
{
	scanf("%d%d%s",&n,&n,&s);
	for(int i=1;i<=n;++i)scanf("%s",&q[i]),AC.insert(q[i]);
	AC.bfs(),AC.bfs2(s);
	for(int i=1;i<=n;++i)printf("%d\n",AC.ask(q[i]));
	return 0;
}