题目
题解
方法一:trie树
标签是AC自动机,但明显不是AC自动机啊,根本没有失配指针的过程,只能叫Tire树吧
我们对每一个单词建立好Tire树之后,在树上进行bfs,进行匹配
代码
#include <cstdio>
#include <cstring>
#include <bitset>
using namespace std;
const int Maxn = 0x3f3f3f3f;
const int N = 505, L = N << 1, M = L * 350;
int T = 1, n, Ans, len, t = 0, w = 1;
int G[M][4], val[M], Q[M][2];
char s[L], a[N], c[255];
bitset<L> vis[M];
inline Ins()
{
scanf("%s", a + 1); int l = strlen(a + 1), x = 1;
for (int i = 1; i <= l; ++i)
{
int y = c[a[i]];
if (!G[x][y]) G[x][y] = ++T;
x = G[x][y];
}
val[x]++;
}
inline void Push(const int x, const int stp)
{
if (!x || stp > len) return;
if (!vis[x][stp])
{
vis[x][stp] = 1; w = (w + 1) % M;
Q[w][0] = x; Q[w][1] = stp;
}
}
int main()
{
c['A'] = 0; c['C'] = 1; c['T'] = 2; c['G'] = 3;
c['?'] = -1; c['*'] = -2;
scanf("%s", s + 1); len = strlen(s + 1);
s[0] = '?';
scanf("%d",&n);
for (int i = 1; i <= n; ++i) Ins();
int u, v, I;
Q[1][0] = 1; Q[1][1] = 0; vis[1][0] = 1;
while (t != w)
{
t = (t + 1) % M;
u = Q[t][0]; v = Q[t][1];//0字典树上的位置 1序列中的位置
if (s[v] == '*')
for (int i = 0; i < 4; ++i) Push(G[u][i], v);
if (v == len) {Ans += val[u]; val[u] = 0; continue;}
if (c[s[v + 1]] >= 0) Push(G[u][c[s[v + 1]]], v + 1);
else
{
for (int i = 0; i < 4; ++i) Push(G[u][i], v + 1);
if (s[v + 1] == '*') Push(u, v + 1);
}
}
printf("%d\n",n-Ans);
return 0;
}
/*
A*G?C
3
AGTC
AGTGTC
AGTGC
*/