BZOJ 4327 [JSOI2012]玄武密码 (AC自动机)

题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=4327

题解:

做法挺显然，建出AC自动机之后在上面跑，标记所有走过的点，然后再进行递推，如果\(fail[x]\)被标记则\(x\)被标记，然后每一个关键点往上找即可

有一个常数优化的技巧: 如果枚举每个\(x\)再vis[fail[x]]|=vis[x]的话好像还需要按深度排序，不如直接每到一个点直接不停地跳fail一路标记，直到遇到已标记点为止

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;

const int N = 1e5;
const int LEN = 100;
const int M = 1e7;
const int SIZ = 1e7;
const int S = 4;
char b[N+3][LEN+3];
char a[M+3];
int len[N+3];
int son[SIZ+3][S+3];
int fail[SIZ+3];
int id[SIZ+3];
bool vis[SIZ+3];
int que[SIZ+3];
int ans[N+3];
int n,m,siz;

int decode(char x)
{
	if(x=='E') return 1;
	else if(x=='S') return 2;
	else if(x=='W') return 3;
	else if(x=='N') return 4;
}

void insertstr(int sid)
{
	int u = 0;
	for(int i=1; i<=len[sid]; i++)
	{
		if(son[u][b[sid][i]]==0) {siz++; son[u][b[sid][i]] = siz;}
		u = son[u][b[sid][i]];
	}
	id[u] = sid;
}

void buildACA()
{
	int head = 1,tail = 0;
	for(int i=1; i<=S; i++)
	{
		if(son[0][i]) {tail++; que[tail] = son[0][i];} fail[son[0][i]] = 0;
	}
	while(head<=tail)
	{
		int u = que[head]; head++;
		for(int i=1; i<=S; i++)
		{
			if(son[u][i]) {fail[son[u][i]] = son[fail[u]][i]; tail++; que[tail] = son[u][i];}
			else {son[u][i] = son[fail[u]][i];}
		}
	}
}

int main()
{
	scanf("%d%d",&m,&n);
	scanf("%s",a+1); for(int i=1; i<=m; i++) a[i] = decode(a[i]);
	for(int i=1; i<=n; i++)
	{
		scanf("%s",b[i]+1); len[i] = strlen(b[i]+1);
		for(int j=1; j<=len[i]; j++) b[i][j] = decode(b[i][j]);
		insertstr(i);
	}
	buildACA();
	int u = 0; vis[0] = true;
	for(int i=1; i<=m; i++)
	{
		vis[u] = true;
		u = son[u][a[i]];
		for(int j=u; vis[j]==false; j=fail[j]) {vis[j] = true;}
	}
	for(int i=1; i<=n; i++)
	{
		u = 0;
		for(int j=1; j<=len[i]; j++)
		{
			u = son[u][b[i][j]];
			if(!vis[u]) {break;}
			ans[i]++;
		}
	}
	for(int i=1; i<=n; i++) printf("%d\n",ans[i]);
	return 0;
}