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目录

• • 算法能解决的问题
  • 算法原理
  • 模板代码实现
  • Trier图
  • 例题
  • • AC自动机朴素写法
    • Trie图 + 路径优化写法

算法能解决的问题

给定$n$个模式串, 给定主串$s$, 非常高的效率检查主串$s$中是否存在某个模式串

算法原理

AC自动机本质是在Trie树上用KMP算法的思想的结果
假设在Trie插入如下模式串she, he, say, shr, her

将KMP算法的ne数组类比到AC自动机中

假设考虑当前子串she, 节点$e$存储的是以$e$结尾的某个后缀(后缀有e, he), 和某个最长真前缀匹配的最长前缀的结尾下标

在上述Trie树中, she的最长匹配真前缀在$he$的结尾

上图是样例构建出AC自动机后的情况

问题就变成了如何计算这些指针?
因为KMP算法计算ne数组的过程是根据$[0,i-1]$的ne值递推出来的

类比于KMP算法计算ne数组的过程

• 计算AC自动机的fail数组, 用$bfs$搜索, 将$u$视为上一层, 当前层是$tr[u][i]$
• j = fail[u]就类比于KMP算法的j = ne[i - 1]
• 然后向前找, 如果失败j = fail[j], 类比到KMP算法, j = ne[j - 1]或者j = ne[j], 两种写法都是对的, 只不过初始下标不同
• 如果找到了匹配的位置if (tr[j][i]), 那么j = tr[j][i], 类比到KMP算法if (s[j] == s[i]) j++或者if (s[j + 1] == s[i]) j++
• 然后记录当前fail值, fail[c] = j, 类比于KMP算法ne[i] = j

伪代码如下
q是队列, h是队头, t是队尾, tr是Trie树, c是当前位置i, u是上一层位置i - 1, j = fail[u]

因为对于Trie树来说, 节点$0$是根节点, 因此插入的时候是从节点$1$开始的, 因此AC自动机向前寻找fail指针的过程是j = fail[j], 而不是j = fail[j - 1]

while (h <= t) {
// 类比为i - 1
int u = q[h++];
for (int i = 0; i < 26; ++i) {
int c = tr[u][i];
// j = ne[i - 1]
int j = fail[u];
while (j && !tr[j][i]) j = fail[j];
if (tr[j][i]) j = tr[j][i];
fail[c] = j;
q[++t] = c;
}
}

类比于KMP算法, AC自动机的匹配方式和构建AC自动机类似

模板代码实现

使用cnt[p]记录如果当前位置的节点编号是单词结尾, 将该位置$+1$, 因此在统计单次数量的时候, 需要将指针$p$能到达的字符串都遍历一遍, 具体的来说

假设当前指针指向了she的结尾e, $p$指针需要移动到右上方, 将he的cnt值进行累加, 因为she出现了, he也必定出现

代码实现

#include <bits/stdc++.h>using namespace std;const int N = 1e4 + 10, M = 1e6 + 10, S = 55;int n;int tr[N * S][26], idx, cnt[N * S];int fail[N * S];int q[N * S], h, t;void insert(string &s) {int p = 0;for (int i = 0; s[i]; ++i) {int c = s[i] - 'a';if (!tr[p][c]) tr[p][c] = ++idx;p = tr[p][c];}cnt[p]++;}void build() {h = 0, t = -1;for (int i = 0; i < 26; ++i) {if (tr[0][i]) q[++t] = tr[0][i];}while (h <= t) {int u = q[h++];for (int i = 0; i < 26; ++i) {int v = tr[u][i];if (!v) continue;int j = fail[u];while (j && !tr[j][i]) j = fail[j];if (tr[j][i]) j = tr[j][i];fail[v] = j;q[++t] = v;}}}void solve() {memset(tr, 0, sizeof tr);memset(cnt, 0, sizeof cnt);memset(fail, 0, sizeof fail);idx = 0;cin >> n;for (int i = 0; i < n; ++i) {string p;cin >> p;insert(p);}build();string s;cin >> s;int ans = 0, j = 0;// 注意在计算主串的时候子节点下标是c = s[i] - 'a';for (int i = 0; s[i]; ++i) {int c = s[i] - 'a';while (j && !tr[j][c]) j = fail[j];if (tr[j][c]) j = tr[j][c];int p = j;while (p) {ans += cnt[p];cnt[p] = 0;p = fail[p];}}cout << ans << '\n';}int main() {ios::sync_with_stdio(false);cin.tie(0);int T;cin >> T;while (T--) solve();return 0;}

Trier图

将代码的向前寻找fail指针的过程进行优化, 具体的来说

对于当前儿子$v$

• 如果当前儿子(是父节点的第$i$个儿子)不存在, 那么当前儿子节点指向父节点的失败指针的第$i$个儿子上
• 如果当前儿子存在, 将当前儿子节点的失败指针指向父节点的失败指针的第$i$个儿子上, 然后将当前节点$v$入队

因为所有的失败指针都指向某个最终节点, 因此在查询过程中, 可以直接查询, 具体的来说

这一部分, 可以直接写成j = tr[j][c]
代码优化后

#include <bits/stdc++.h>using namespace std;const int N = 1e4 + 10, M = 1e6 + 10, S = 55;int n;int tr[N * S][26], idx, cnt[N * S];int fail[N * S];int q[N * S], h, t;void insert(string &s) {int p = 0;for (int i = 0; s[i]; ++i) {int c = s[i] - 'a';if (!tr[p][c]) tr[p][c] = ++idx;p = tr[p][c];}cnt[p]++;}void build() {h = 0, t = -1;for (int i = 0; i < 26; ++i) {if (tr[0][i]) q[++t] = tr[0][i];}while (h <= t) {int u = q[h++];for (int i = 0; i < 26; ++i) {int v = tr[u][i];if (!v) tr[u][i] = tr[fail[u]][i];else {fail[v] = tr[fail[u]][i];q[++t] = v;}}}}void solve() {memset(tr, 0, sizeof tr);memset(cnt, 0, sizeof cnt);memset(fail, 0, sizeof fail);idx = 0;cin >> n;for (int i = 0; i < n; ++i) {string p;cin >> p;insert(p);}build();string s;cin >> s;int ans = 0, j = 0;// 注意在计算主串的时候子节点下标是c = s[i] - 'a';for (int i = 0; s[i]; ++i) {int c = s[i] - 'a';j = tr[j][c];int p = j;while (p) {ans += cnt[p];cnt[p] = 0;p = fail[p];}}cout << ans << '\n';}int main() {ios::sync_with_stdio(false);cin.tie(0);int T;cin >> T;while (T--) solve();return 0;}

例题

定义状态表示$f(i,j)$表示考虑生成前$i$个字符, 并且当前指向了AC自动机的第$j$个位置, 的所有方案中需要改变的字符数量最少的值

AC自动机朴素写法

#include <bits/stdc++.h>using namespace std;const int N = 55, M = 1010, K = 30, INF = 0x3f3f3f3f;int T, n, m;int tr[N * K][4], idx, cnt[N * K];string s;int q[N * K], h, t;int fail[N * K];int f[M][N * K];int get(char c) {if (c == 'A') return 0;else if (c == 'G') return 1;else if (c == 'C') return 2;else return 3;}void insert(string &s) {int p = 0;for (int i = 0; s[i]; ++i) {int t = get(s[i]);if (!tr[p][t]) tr[p][t] = ++idx;p = tr[p][t];}cnt[p] = 1;}void build() {h = 0, t = -1;for (int i = 0; i < 4; ++i) {if (tr[0][i]) q[++t] = tr[0][i];}while (h <= t) {int u = q[h++];for (int i = 0; i < 4; ++i) {int v = tr[u][i];if (!v) continue;int j = fail[u];while (j && !tr[j][i]) j = fail[j];if (tr[j][i]) j = tr[j][i];fail[v] = j;q[++t] = v;}}}void solve() {memset(tr, 0, sizeof tr);memset(cnt, 0, sizeof cnt);memset(fail, 0, sizeof fail);memset(f, 0x3f, sizeof f);idx = 0;f[0][0] = 0;for (int i = 0; i < n; ++i) {string s;cin >> s;insert(s);}build();cin >> s;m = s.size();s = ' ' + s;for (int i = 0; i < m; ++i) {for (int j = 0; j <= idx; ++j) {for (int k = 0; k < 4; ++k) {int cost = get(s[i + 1]) != k;int p = j;while (p && !tr[p][k]) p = fail[p];if (tr[p][k]) p = tr[p][k];else p = 0;// 关键点, 检查在后缀匹配的前缀中是否有某个前缀是致病片段int x = p;bool flag = true;while (x) {if (cnt[x]) {flag = false;break;}x = fail[x];}if (flag) f[i + 1][p] = min(f[i + 1][p], f[i][j] + cost);}}}int ans = INF;for (int i = 0; i <= idx; ++i) ans = min(ans, f[m][i]);if (ans == INF) ans = -1;printf("Case %d: %d\n", ++T, ans);}int main() {ios::sync_with_stdio(false);cin.tie(0);while (cin >> n, n) solve();return 0;}

Trie图 + 路径优化写法

如果当前后缀所匹配的某个前缀有致病片段, 因为是匹配的, 当前后缀也包含那个前缀, 因此当前后缀也有致病片段, 因此可以做cnt[v] |= cnt[fail[v]]这样的优化

#include <bits/stdc++.h>using namespace std;const int N = 55, M = 1010, K = 30, INF = 0x3f3f3f3f;int T, n, m;int tr[N * K][4], idx, cnt[N * K];string s;int q[N * K], h, t;int fail[N * K];int f[M][N * K];int get(char c) {if (c == 'A') return 0;else if (c == 'G') return 1;else if (c == 'C') return 2;else return 3;}void insert(string &s) {int p = 0;for (int i = 0; s[i]; ++i) {int t = get(s[i]);if (!tr[p][t]) tr[p][t] = ++idx;p = tr[p][t];}cnt[p] = 1;}void build() {h = 0, t = -1;for (int i = 0; i < 4; ++i) {if (tr[0][i]) q[++t] = tr[0][i];}while (h <= t) {int u = q[h++];for (int i = 0; i < 4; ++i) {int v = tr[u][i];if (!v) tr[u][i] = tr[fail[u]][i];else {fail[v] = tr[fail[u]][i];// 能够跳转到的某个前缀是否含有致病片段, 如果某个前缀有, 当前位置也有cnt[v] |= cnt[fail[v]];q[++t] = v;}}}}void solve() {memset(tr, 0, sizeof tr);memset(cnt, 0, sizeof cnt);memset(fail, 0, sizeof fail);memset(f, 0x3f, sizeof f);idx = 0;f[0][0] = 0;for (int i = 0; i < n; ++i) {string s;cin >> s;insert(s);}build();cin >> s;m = s.size();s = ' ' + s;for (int i = 0; i < m; ++i) {for (int j = 0; j <= idx; ++j) {for (int k = 0; k < 4; ++k) {int cost = get(s[i + 1]) != k;int p = tr[j][k];if (!cnt[p]) f[i + 1][p] = min(f[i + 1][p], f[i][j] + cost);}}}int ans = INF;for (int i = 0; i <= idx; ++i) ans = min(ans, f[m][i]);if (ans == INF) ans = -1;printf("Case %d: %d\n", ++T, ans);}int main() {ios::sync_with_stdio(false);cin.tie(0);while (cin >> n, n) solve();return 0;}