String

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题目链接

题意

给你若干个字符串，每次询问一个$S$,$T$分别作为前缀和后缀不重叠的字符串的数量有多少。

思路

直接找不好找，我们可以这样建AC自动机：
每个原串变成$str+$’#’$+str$，每个前缀后缀变成‘#’插入AC自动机，这样我们查询就可以满足前缀和后缀。
还有一个条件，要求不能重叠，那我们再加上一个长度限制就好了。

Code

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50

const int N = 5e5 + 5;
const int maxn = N * 30;
int root, sz, ch[maxn][27], fail[maxn], ac_len[maxn], ed[maxn];
const int Q = 1e5 + 5;
int pos[Q];
int react(char c)
{
	if (c >= 'a' && c <= 'z')
		return c - 'a';
	return 26;
}

void init()
{
	root = sz = 0;
	for (int c = 0;c <= 26; c++)
		ch[root][c] = 0;
	ac_len[root] = 0;
	ed[root] = 0;
}

int newnode()
{
	sz++;
	for (int c = 0; c <= 26; c++)
		ch[sz][c] = 0;
	ed[sz] = 0;
	return sz;
}

void insert(char buf[], int id)
{
	int len = strlen(buf);
	int now = root;
	for (int i = 0; i < len; i++)
	{
		int c = react(buf[i]);
		if (ch[now][c] == 0)
		{
			ch[now][c] = newnode();
			ac_len[ch[now][c]] = i + 1;
		}
		now = ch[now][c];
	}
	pos[id] = now;
//	dbg(id, pos[id]);
}

void get_fail()
{
	queue<int> q;
	for (int c = 0; c <= 26; c++)
	{
		if (ch[root][c] == 0)
			ch[root][c] = root;
		else
		{
			int u = ch[root][c];
			fail[u] = root;
			q.push(u);
		}
	}
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
//		dbg(u);
		for (int c = 0; c <= 26; c++)
		{
			if (ch[u][c] == 0)
			{
				ch[u][c] = ch[fail[u]][c];
				continue;
			}
			int v = ch[u][c];
			fail[v] = ch[fail[u]][c];
			q.push(v);
		}
	}
	//puts("over2");
}

void query(char buf[], int Len)
{
	int len = strlen(buf);
	int now = root;
//	puts(buf);
//	dbg(Len);
	for (int i = 0; i < len; i++)
	{
		int c = react(buf[i]);
		while (now != root && !ch[now][c])
			now = fail[now];
		now = ch[now][c];
		int tmp = now;
		while (tmp != root)
		{
//			printf("-----------------%d, ac_len %d\n", tmp, ac_len[tmp]);
			if (Len >= ac_len[tmp])
				ed[tmp]++;
			tmp = fail[tmp];
		}
	}
}
char s[N];
char tmp[N * 2];
int len[Q];
char pre[N];
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int n, q;
		scanf("%d%d", &n, &q);
		int cpos = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%s", s + cpos);
			len[i] = strlen(s + cpos);
			cpos += len[i];
		}
		init();
		for (int k = 1; k <= q; k++)
		{
			scanf("%s", pre);
			scanf("%s", tmp);
			int cur = strlen(tmp);
			tmp[cur++] = '#';
			int j = 0;
			while (pre[j])
			{
				tmp[cur++] = pre[j++];
			}
			tmp[cur] = 0;
//			puts(tmp);
			insert(tmp, k);
		}
	//	puts("build over");
		get_fail();
		cpos = 0;
	//	puts("over");
		for (int i = 1; i <= n; i++)
		{
			int j;
			for (j = 0; j < len[i]; j++)
				tmp[j] = tmp[j + len[i] + 1] = s[cpos++];
			tmp[j] = '#';
			tmp[j + len[i] + 1] = 0;
//			puts(tmp);
			query(tmp, len[i] + 1);
		}
		for (int i = 1; i <= q; i++)
		{
			printf("%d\n", ed[pos[i]]);
		}
	}
	return 0;
}
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