hdoj 2222 Keywords Search

題目大意：給出一組單詞，和一個文本，要你統計有多少個單詞出現在這個文本中。

解題思路：簡單的AC自動機入門，先把單詞建立單詞樹，注意單詞有重複出現，然後建構失敗指針。

最後用文本在自動機上做查詢統計個數

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;

const int maxn = 55;
const int maxm = 1000010;
const int kind = 26;

struct node
{
	node *fail;
	node *next[kind];
	int isLeaf;
	node()
	{
		fail = NULL;
		isLeaf = 0;
		memset(next, 0, sizeof(next));
	}
};

int t, n;
char str[maxn], text[maxm];

void insert(char *s, node *root);
void build(node *root);
int query(char *s, node *root);

int main()
{
	scanf("%d", &t);
	while(t-- != 0)
	{
		scanf("%d", &n);
		node *root = new node;
		for(int i = 0; i < n; i++)
		{
			scanf("%s", str);
			insert(str, root);
		}
		build(root);
		scanf("%s", text);
		printf("%d\n", query(text, root));
	}
	
	return 0;
}

void insert(char *s, node *root)
{
	int i = 0, index;
	node *p = root;
	while(s[i] != '\0')
	{
		index = s[i] - 'a';
		if(!p->next[index])
			p->next[index] = new node();
		p = p->next[index];
		i++;
	}
	p->isLeaf++;
}

void build(node *root)
{
	node *tmp , *p;
	queue<node *> que;
	root->fail = NULL;
	que.push(root);
	while(!que.empty())
	{
		tmp = que.front();
		p = NULL;
		que.pop();
		for(int i = 0; i < kind; i++)
		{
			if(tmp->next[i] != NULL)
			{
				if(tmp == root)
					tmp->next[i]->fail = root;
				else
				{
					p = tmp->fail;
					while(p != NULL)
					{
						if(p->next[i] != NULL)
						{
							tmp->next[i]->fail = p->next[i];
							break;
						}
						p = p->fail;
					}
					if(p == NULL)
						tmp->next[i]->fail = root;
				}
				que.push(tmp->next[i]);
			}
		}
	}
}

int query(char *s, node *root)
{
	int ans = 0, i = 0, index;
	node *p = root, *tmp = NULL;
	while(s[i] != '\0')
	{
		index = s[i] - 'a';
		while(p != NULL && p->next[index] == NULL)
			p = p->fail;
		if(p == NULL)
			p = root;
		else
			p = p->next[index];
		tmp = p;
		while(tmp != NULL && tmp->isLeaf != 0)
		{
			ans += tmp->isLeaf;
			tmp->isLeaf = 0; 
			tmp = tmp->fail;
		}
		i++;
	}
	return ans;
}