好久以前做的AC自动机题目了，趁还没忘掉赶紧回忆一下
AC自动机简单说就是预先知道所有的模式串，建立Trie树来进行目标串的匹配
【模板】AC自动机（二次加强版）
洛谷P5357
AC自动机裸题，上了个last优化过去了= =
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
//clock_t c1 = clock();
//std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 2e6 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
int tree[MAXN][26]; //tree[i][j]表示节点i的第j个儿子的节点编号
int flagg[MAXN]; //表示以该节点结尾是一个单词
// int cntword[MAXN]; //表示以该结点为结尾的单词数量
int sum[MAXN];
int fail[MAXN]; //fail指针
int vis[MAXN];
int mp[MAXN];
int tot; //总节点数
int last[MAXN];
void insert_(char *str, int num)
{
int len = strlen(str);
int root = 0;
for (int i = 0; i < len; i++)
{
int id = str[i] - 'a';
if (!tree[root][id])
tree[root][id] = ++tot;
sum[tree[root][id]]++;
root = tree[root][id];
}
if (!flagg[root])
flagg[root] = num;
vis[num] = root;
}
void init() //最后清空，节省时间
{
for (int i = 0; i <= tot; i++)
{
sum[i] = 0;
flagg[i] = false;
for (int j = 0; j < 26; j++)
tree[i][j] = 0;
last[i] = 0;
}
tot = 0; //RE有可能是这里的问题
}
void build()
{
queue<int> q;
for (int i = 0; i < 26; i++)
{
if (tree[0][i]) //如果第二层存在该结点，则将结点压入队列，fail指针指向根结点
{
fail[tree[0][i]] = 0;
q.push(tree[0][i]);
}
}
while (!q.empty())
{
int now = q.front();
q.pop();
for (int i = 0; i < 26; i++)
{
if (tree[now][i]) //存在该结点，该结点的fail指针指向父节点的fail指针指向的结点的子结点
{
fail[tree[now][i]] = tree[fail[now]][i];
q.push(tree[now][i]);
last[tree[now][i]] = flagg[fail[tree[now][i]]] ? fail[tree[now][i]] : last[fail[tree[now][i]]];
}
else
tree[now][i] = tree[fail[now]][i]; //不存在该结点，直接让该子结点指向父节点的fail指针指向的结点的子结点
}
}
}
void query(char *s)
{
int len = strlen(s);
int now = 0, ans = 0;
for (int i = 0; i < len; i++)
{
now = tree[now][s[i] - 'a'];
if (flagg[now])
{
mp[now]++;
// flagg[now] = 0;
}
int t = now;
while (last[t])
{
t = last[t];
if (flagg[t])
{
mp[t]++;
// flagg[t] = 0;
}
}
}
}
char s[MAXN]; //模式串
char t[MAXN]; //目标串
int main()
{
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf(" %s", s);
insert_(s, i);
}
build();
scanf(" %s", t);
query(t);
for (int i = 1; i <= n; i++)
printf("%d\n", mp[vis[i]]);
return 0;
}HDU-1277全文检索
#include <bits/stdc++.h>
using namespace std;
/* freopen("k.in", "r", stdin);
freopen("k.out", "w", stdout); */
//clock_t c1 = clock();
//std::cerr << "Time:" << clock() - c1 <<"ms" << std::endl;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define de(a) cout << #a << " = " << a << endl
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef vector<int, int> VII;
#define inf 0x3f3f3f3f
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll MAXN = 1e6 + 7;
const ll MAXM = 1e6 + 7;
const ll MOD = 1e9 + 7;
const double eps = 1e-6;
const double pi = acos(-1.0);
int tree[MAXN][15]; //tree[i][j]表示节点i的第j个儿子的节点编号
bool flagg[MAXN]; //表示以该节点结尾是一个单词
int cntword[MAXN]; //表示以该结点为结尾的单词数量
int mark[MAXN]; //记录以该结点为结尾的单词的原来编号
int sum[MAXN];
int fail[MAXN]; //fail指针
int tot; //总节点数
void insert_(char *str, int id)
{
int len = strlen(str);
int root = 0;
for (int i = 0; i < len; i++)
{
int id = str[i] - '0';
if (!tree[root][id])
tree[root][id] = ++tot;
sum[tree[root][id]]++;
root = tree[root][id];
}
flagg[root] = true;
cntword[root]++;
mark[root] = id;
}
void init() //最后清空，节省时间
{
for (int i = 0; i <= tot; i++)
{
sum[i] = 0;
flagg[i] = false;
cntword[i] = -1;
for (int j = 0; j < 10; j++)
tree[i][j] = 0;
}
tot = 0; //RE有可能是这里的问题
}
void build()
{
queue<int> q;
for (int i = 0; i < 10; i++)
{
if (tree[0][i]) //如果第二层存在该结点，则将结点压入队列，fail指针指向根结点
{
fail[tree[0][i]] = 0;
q.push(tree[0][i]);
}
}
while (!q.empty())
{
int now = q.front();
q.pop();
for (int i = 0; i < 10; i++)
{
if (tree[now][i]) //存在该结点，该结点的fail指针指向父节点的fail指针指向的结点的子结点
{
fail[tree[now][i]] = tree[fail[now]][i];
q.push(tree[now][i]);
}
else
tree[now][i] = tree[fail[now]][i]; //不存在该结点，直接让该子结点指向父节点的fail指针指向的结点的子结点
}
}
}
bool flag = true;
void query(char *s)
{
int len = strlen(s);
int now = 0;
for (int i = 0; i < len; i++)
{
now = tree[now][s[i] - '0'];
for (int j = now; j && ~cntword[j]; j = fail[j])
{
if (!flag)
printf(" ");
printf("[Key No. %d]", mark[j]);
flag = false;
cntword[j] = -1;
}
}
}
char s[MAXN]; //模式串
char t[MAXN]; //目标串
char temp[MAXN];
int main()
{
memset(cntword, -1, sizeof(cntword));
int n, m;
while (~scanf("%d%d", &n, &m))
{
for (int i = 0; i < n; i++)
{
scanf(" %s", temp);
if (!i)
strcpy(t, temp);
else
strcat(t, temp);
}
for (int j = 1; j <= m; j++)
{
int num;
scanf("%*s%*s%d] %s", &num, s);
insert_(s, num);
}
build();
printf("Found key: ");
query(t);
puts("");
init();
}
return 0;
}