UESTC 8 God Only Knows!

最新推荐文章于 2020-10-09 13:24:49 发布

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本文介绍了一个使用AC自动机解决寻找不包含特定病毒子串的问题。通过构建AC自动机，可以高效地处理多个模式匹配任务。文章详细解释了输入输出格式、样例以及AC自动机的具体实现。

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Description

Zplinti1 is given a big and long string , together with a list of strings that are viruses. He wants to find the number of substrings of , so that it does not contain any viruses. Same substrings with different starting positions are regarded as different!

Zplinti1 finds this problem difficult enough, he thinks Only God Can Solve This Problem, but you don’t think so, right?

Input

The first line of input contains a number , indicating the number of cases. ()
For each case, the first line is a string , with length no more than . The second line is a number , which is the number of virus strings. Then lines comes, each with a string.

All the strings contains lowercase letters from a to z only. The total length of all virus strings in one case will be no more than .

Output

For each case, output Case #i: first. ( is the number of the test case, from to ). Then output the number of substrings of that do not contain any virus.

Sample Input

3
aabc
0
aabbaa
1
a
abcdefg
2
bcd
ef

Sample Output

Case #1: 10
Case #2: 3
Case #3: 14

Hint

A substring of a string is a continuous string taken from . For example if we take out the letters from the to the of the string, then the substring will be ().

The original string is a substring of itself.

When we say string contains string , it means that is a substring of .

Source

The 11th UESTC Programming Contest Final

ac自动机的应用

#include<stdio.h>
#include<string.h>
#include<queue>
#include<malloc.h>
using namespace std;
const int maxn = 1000005;
int T, n, t = 0;
long long ans;

class tire
{
public:
	tire *down[26], *next;
	int end;
	tire(){ next = NULL; end = 0; memset(down, 0, sizeof(down)); }
	void assign(){ next = NULL; end = 0; memset(down, 0, sizeof(down)); }
};

class ac_automaton
{	
private:
	tire *root;
	char s[maxn];
	char S[maxn];
public:
	ac_automaton(){ root = new tire; }
	void insert()
	{
		scanf("%s",s);
		tire* j = root;
		for (int i = 0, k; s[i]; i++)
		{
			k = s[i] - 'a';
			if (!j->down[k]) j->down[k] = new tire;
			j = j->down[k];
		}
		j->end = strlen(s);
	}

	void getnext()
	{
		queue<tire*> p;
		tire *q, *k, *j;

		root->next = root;
		for (int i = 0; i < 26; i++)
		if (root->down[i])
		{
			q = root->down[i];
			q->next = root;
			p.push(q);
		}
		else root->down[i] = root;

		while (!p.empty())
		{
			q = p.front();    p.pop();
			k = q->next;

			for (int i = 0; i < 26; i++)
			if (q->down[i])
			{
				j = q->down[i];
				if (k->down[i]->end)
				{
					if (j->end) j->end = min(k->down[i]->end, j->end);
					else j->end = k->down[i]->end;
				}
				j->next = k->down[i];
				p.push(q->down[i]);
			}
			else q->down[i] = k->down[i];
		}
	}	

	int getmother() { scanf("%s", S); root = new tire; return strlen(S); }

	void work_out()
	{
		getnext();
		tire *j = root, *u;
		for (int i = 0, k, u = ans = 0; S[i]; i++)
		{
			k = S[i] - 'a';
			j = j->down[k];
			
			if (j->end)
			{
				u = min(u + 1, j->end - 1);
				ans += u;
			}
			else { u++; ans += u; }
		}
		printf("Case #%d: %lld\n", ++t, ans);
	}
};

ac_automaton f;

int main()
{
	scanf("%d", &T);	
	while (T--)
	{
		int u = f.getmother();
		scanf("%d", &n);
		while (n--) f.insert();
		f.work_out();
	}
	return 0;
}