【题目链接】
【思路要点】
- 对模式串的集合建立AC自动机,让主串在上面匹配,每遇到一个模式串的末尾便将其删去即可。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); }
template <typename T> void read(T &x) {
x = 0; int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
template <typename T> void write(T x) {
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
write(x);
puts("");
}
struct ACAutomaton {
struct Node {
int child[26];
int fail, cnt;
} a[MAXN];
int root, size;
void insert(char *s) {
int now = root;
int len = strlen(s + 1);
for (int i = 1; i <= len; i++) {
int tmp = s[i] - 'a';
if (a[now].child[tmp] == 0) a[now].child[tmp] = ++size;
now = a[now].child[tmp];
}
a[now].cnt = len;
}
void init() {
static int q[MAXN];
int l = 0, r = -1;
for (int i = 0; i < 26; i++)
if (a[root].child[i]) {
q[++r] = a[root].child[i];
a[a[root].child[i]].fail = root;
}
while (l <= r) {
int tmp = q[l++];
for (int i = 0; i < 26; i++)
if (a[tmp].child[i]) {
q[++r] = a[tmp].child[i];
a[a[tmp].child[i]].fail = a[a[tmp].fail].child[i];
} else a[tmp].child[i] = a[a[tmp].fail].child[i];
}
}
void calc(char *s) {
int len = strlen(s + 1);
static int pos[MAXN];
static char ans[MAXN];
int anslen = 0;
for (int i = 1; i <= len; i++) {
int tmp = s[i] - 'a';
anslen++;
ans[anslen] = s[i];
pos[anslen] = a[pos[anslen - 1]].child[tmp];
anslen -= a[pos[anslen]].cnt;
}
for (int i = 1; i <= anslen; i++)
printf("%c", ans[i]);
}
} ACAM;
char s[MAXN], t[MAXN];
int main() {
scanf("\n%s", s + 1);
int n; read(n);
for (int i = 1; i <= n; i++) {
scanf("\n%s", t + 1);
ACAM.insert(t);
}
ACAM.init();
ACAM.calc(s);
return 0;
}