AC自动机板子
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例题:HDU - 2222
给\(n\)个字符串,一个模式串。然后输出匹配次数。
代码
#include <bits/stdc++.h>
#define FOPI freopen("in.txt", "r", stdin)
#define FOPO freopen("out.txt", "w", stdout)
#define FOR(i,x,y) for (int i = x; i <= y; i++)
#define ROF(i,x,y) for (int i = x; i >= y; i--)
using namespace std;
typedef long long LL;
struct Trie
{
int next[500010][26], fail[500010], end[500010];
int root, L;
int newnode()
{
FOR(i, 0, 26-1) next[L][i] = -1;
end[L++] = 0;
return L-1;
}
void init() { L = 0; root = newnode(); }
void insert(char buf[])
{
int len = strlen(buf), now = root;
FOR(i, 0, len-1)
{
if (next[now][buf[i] - 'a'] == -1)
next[now][buf[i] - 'a'] = newnode();
now = next[now][buf[i] - 'a'];
}
end[now]++;
}
void build()
{
queue<int> Q;
fail[root] = root;
FOR(i, 0, 26-1)
if (next[root][i] == -1) next[root][i] = root;
else
{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
while(!Q.empty())
{
int now = Q.front();
Q.pop();
FOR(i, 0, 26-1)
if (next[now][i] == -1)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]] = next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
int query(char buf[])
{
int len = strlen(buf);
int now = root;
int res = 0;
FOR(i, 0, len-1)
{
now = next[now][buf[i]-'a'];
int temp = now;
while(temp != root)
{
res += end[temp];
end[temp] = 0;
temp = fail[temp];
}
}
return res;
}
// void debug()
// {
// FOR(i, 0, L-1)
// {
// printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], end[i]);
// FOR(j, 0, 26-1) printf("%2d", next[i][j]);
// printf("]\n");
// }
// }
};
char buf[1000010];
Trie ac;
int t, n;
int main()
{
scanf("%d", &t);
FOR(ca, 1, t)
{
scanf("%d", &n);
ac.init();
FOR(i, 0, n-1)
{
scanf("%s", buf);
ac.insert(buf);
}
ac.build();
scanf("%s", buf);
printf("%d\n", ac.query(buf));
}
}
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