AC自动机
AC自动机
感觉非常合理，但是又不会证明，就先这样吧
感觉还有至少三个问题：
1、为什么要在空的子结点上连自己的fail结点的对应子结点（注释1）可以把整个自动机变成一个\(DAG\)
2、为什么u结点不需要参与转移（注释2）同上
3、为什么遇到end == -1就可以break（注释3）这题只需要统计是否出现过不用统计次数
昨天又做了几题，差不多会用板子了（只能意会）
#include<bits/stdc++.h>
using namespace std;
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;
const int INF = 0X3f3f3f3f, N = 1e6 + 10, MOD = 1e9 + 7;
const double eps = 1e-7, pi = acos(-1);
int n;
string s, str;
struct Trie {
int son[26];
int end;
int fail;
} tr[N];
int idx;
void insert(string &a) {
int u = 0;
rep (i, 0, a.length() - 1) {
int v = a[i] - 'a';
if (tr[u].son[v]) u = tr[u].son[v];
else tr[u].son[v] = ++idx, u = idx;
}
tr[u].end++;
}
void build() {
queue<int> Q;
rep (i, 0, 25) if (tr[0].son[i]) Q.push(tr[0].son[i]);
while (!Q.empty()) {
int u = Q.front();
Q.pop();
rep (i, 0, 25) {
int v = tr[u].son[i];
if (!v) tr[u].son[i] = tr[tr[u].fail].son[i];
else tr[v].fail = tr[tr[u].fail].son[i], Q.push(v); // 1
}
}
}
void work() {
cin >> n;
rep (i, 1, n) {
cin >> s;
insert(s);
}
build();
cin >> str;
int u = 0, ans = 0;
rep (i, 0, str.length() - 1) {
u = tr[u].son[str[i] - 'a']; // 2
for (int t = u; t && tr[t].end != -1; t = tr[t].fail) { // 3
ans += tr[t].end;
tr[t].end = -1;
}
}
cout << ans << endl;
}
signed main() {
IO
int test = 1;
// cin >> test;
while (test--) {
work();
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;
const int INF = 0X3f3f3f3f, N = 1e4 + 10, MOD = 1e9 + 7;
const double eps = 1e-7, pi = acos(-1);
int n;
string s, str;
struct Trie {
int son[26];
int end;
int fail;
} tr[N * 26];
int idx;
void insert(string &a) {
int u = 0;
rep (i, 0, a.length() - 1) {
int v = a[i] - 'a';
if (tr[u].son[v]) u = tr[u].son[v];
else tr[u].son[v] = ++idx, u = idx;
}
tr[u].end++;
}
void build() {
queue<int> Q;
rep (i, 0, 25) if (tr[0].son[i]) Q.push(tr[0].son[i]);
while (!Q.empty()) {
int u = Q.front();
Q.pop();
rep (i, 0, 25) {
int v = tr[u].son[i];
if (!v) tr[u].son[i] = tr[tr[u].fail].son[i];
else tr[v].fail = tr[tr[u].fail].son[i], Q.push(v);
}
}
}
void work() {
idx = 0;
memset(tr, 0, sizeof tr);
cin >> n;
rep (i, 1, n) {
cin >> s;
insert(s);
}
build();
cin >> str;
int u = 0, ans = 0;
rep (i, 0, str.length() - 1) {
u = tr[u].son[str[i] - 'a'];
for (int t = u; t && tr[t].end != -1; t = tr[t].fail) {
ans += tr[t].end;
tr[t].end = -1;
}
}
cout << ans << endl;
}
signed main() {
IO
int test = 1;
cin >> test;
while (test--) {
work();
}
return 0;
}
P3796 【模板】AC 自动机（加强版）
前面三题都差不多
#include<bits/stdc++.h>
using namespace std;
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;
const int INF = 0X3f3f3f3f, N = 2e5 + 10, MOD = 1e9 + 7;
const double eps = 1e-7, pi = acos(-1);
int n;
int id[210];
string s, str;
vector<int> v;
struct Trie {
int son[26];
int end;
int fail;
int res;
string t;
} tr[N];
int idx;
int insert(string &a) {
int u = 0;
rep (i, 0, a.length() - 1) {
int v = a[i] - 'a';
if (tr[u].son[v]) u = tr[u].son[v];
else tr[u].son[v] = ++idx, u = idx;
}
tr[u].end++;
tr[u].t = a;
return u;
}
void build() {
queue<int> Q;
rep (i, 0, 25) if (tr[0].son[i]) Q.push(tr[0].son[i]);
while (!Q.empty()) {
int u = Q.front();
v.push_back(u);
Q.pop();
rep (i, 0, 25) {
int v = tr[u].son[i];
if (!v) tr[u].son[i] = tr[tr[u].fail].son[i];
else tr[v].fail = tr[tr[u].fail].son[i], Q.push(v);
}
}
}
void work() {
while (cin >> n, n) {
idx = 0;
v.clear();
memset(tr, 0, sizeof tr);
rep (i, 1, n) {
cin >> s;
id[i] = insert(s);
}
build();
cin >> str;
int u = 0;
rep (i, 0, str.length() - 1) {
u = tr[u].son[str[i] - 'a'];
for (int t = u; t; t = tr[t].fail) if(tr[t].end) tr[t].res++;
}
int mx = 0;
vector<string> ans;
rep (i, 1, idx) {
if (tr[i].res > mx) ans.clear(), ans.push_back(tr[i].t), mx = tr[i].res;
else if (tr[i].res == mx) ans.push_back(tr[i].t);
}
cout << mx << endl;
rep (i, 0, ans.size() - 1) cout << ans[i] << endl;
}
}
signed main() {
IO
int test = 1;
// cin >> test;
while (test--) {
work();
}
return 0;
}
Acwing 1285. 单词
直接统计最坏\(O(n^2)\)，我们发现在自动机的trie图上每个结点的出度有且只有一个，并且指向的是拥有相同后缀的单词，所以我们可以在统计完每个结点的单点贡献后，利用拓扑序进行累加，复杂度\(O(n)\)
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;
const int INF = 0X3f3f3f3f, N = 1e6 + 10, MOD = 1e9 + 7;
const double eps = 1e-7, pi = acos(-1);
int n;
int id[210], f[N];
string s, str;
vector<int> v;
struct Trie {
int son[26];
int end;
int fail;
} tr[N];
int idx;
int insert(string &a) {
int u = 0;
rep (i, 0, a.length() - 1) {
int v = a[i] - 'a';
if (tr[u].son[v]) u = tr[u].son[v];
else tr[u].son[v] = ++idx, u = idx;
f[u]++;
}
tr[u].end++;
return u;
}
void build() {
queue<int> Q;
rep (i, 0, 25) if (tr[0].son[i]) Q.push(tr[0].son[i]);
while (!Q.empty()) {
int u = Q.front();
v.push_back(u);
Q.pop();
rep (i, 0, 25) {
int v = tr[u].son[i];
if (!v) tr[u].son[i] = tr[tr[u].fail].son[i];
else tr[v].fail = tr[tr[u].fail].son[i], Q.push(v);
}
}
}
void work() {
cin >> n;
rep (i, 1, n) {
cin >> s;
id[i] = insert(s);
}
build();
rrep (i, idx - 1, 0) f[tr[v[i]].fail] += f[v[i]];
rep (i, 1, n) cout << f[id[i]] << endl;
}
signed main() {
IO
int test = 1;
// cin >> test;
while (test--) {
work();
}
return 0;
}
P5357 【模板】AC 自动机（二次加强版）
跟上一题类似，先遍历统计原串的单点贡献，再按照拓扑序累加
#include<bits/stdc++.h>
using namespace std;
#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;
const int INF = 0X3f3f3f3f, N = 2e5 + 10, MOD = 1e9 + 7;
const double eps = 1e-7, pi = acos(-1);
int n;
int f[N], id[N];
string s, str;
struct Trie {
int son[26];
int end;
int fail;
int res;
} tr[N];
int idx;
vector<int> vc;
int insert(string &a) {
int u = 0;
rep (i, 0, a.length() - 1) {
int v = a[i] - 'a';
if (tr[u].son[v]) u = tr[u].son[v];
else tr[u].son[v] = ++idx, u = idx;
}
tr[u].end++;
return u;
}
void build() {
queue<int> Q;
rep (i, 0, 25) if (tr[0].son[i]) Q.push(tr[0].son[i]);
while (!Q.empty()) {
int u = Q.front();
Q.pop();
vc.push_back(u);
rep (i, 0, 25) {
int v = tr[u].son[i];
if (!v) tr[u].son[i] = tr[tr[u].fail].son[i];
else tr[v].fail = tr[tr[u].fail].son[i], Q.push(v);
}
}
}
void work() {
cin >> n;
rep (i, 1, n) {
cin >> s;
id[i] = insert(s);
}
build();
cin >> str;
int u = 0;
rep (i, 0, str.length() - 1) {
u = tr[u].son[str[i] - 'a'];
f[u]++;
}
rrep (i, idx - 1, 0) f[tr[vc[i]].fail] += f[vc[i]];
rep (i, 1, n) cout << f[id[i]] << endl;
}
signed main() {
IO
int test = 1;
// cin >> test;
while (test--) {
work();
}
return 0;
}
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