LOJ10059 Censoring

题目链接
维护一个栈,表示每个输出字符对应\(tire\)图上的节点,在\(tire\)图上游走时,如果走到某个节点是敏感词结束的地方,就弹出敏感词,并且栈回退到到这个敏感词之前的位置。

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#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 5;
int n, tr[N][26], fail[N], id[N], tot, stk[N], top, o[N];
char s[N], qaq[N];

void insert(const char *s) {
int u = 0, n = strlen(s + 1);
for (int i = 1; i <= n; i++) {
int c = s[i] - 'a';
if (!tr[u][c]) tr[u][c] = ++tot;
u = tr[u][c];
}
id[u] = n;
}
void build() {
queue<int> q;
for (int i = 0; i < 26; i++) if (tr[0][i]) q.push(tr[0][i]);
while (!q.empty()) {
int u = q.front(); q.pop();
for (int i = 0; i < 26; i++) {
if (tr[u][i]) fail[tr[u][i]] = tr[fail[u]][i], q.push(tr[u][i]);
else tr[u][i] = tr[fail[u]][i];
}
}
}

int main() {
scanf("%s", qaq + 1);
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%s", s + 1), insert(s);
build();
int m = strlen(qaq + 1), u = 0;
for (int i = 1; i <= m; i++) {
int c = qaq[i] - 'a';
u = tr[u][c];
stk[++top] = u;
o[top] = i;
if (id[u]) {
top -= id[u];
u = stk[top];
}
}
for (int i = 1; i <= top; i++) printf("%c", qaq[o[i]]);
return 0;
}