AC自动机（二次加强版）

要找每个串的出现次数，实际上就是在fail树上进行处理了，我们知道，在fail树的祖先节点上的点，一定是在这之前的前缀的点，所以直接进行跳转就可以了。
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <string>
  5 #include <cstring>
  6 #include <algorithm>
  7 #include <limits>
  8 #include <vector>
  9 #include <stack>
 10 #include <queue>
 11 #include <set>
 12 #include <map>
 13 #include <bitset>
 14 #include <unordered_map>
 15 #include <unordered_set>
 16 #define lowbit(x) ( x&(-x) )
 17 #define pi 3.141592653589793
 18 #define e 2.718281828459045
 19 #define INF 0x3f3f3f3f
 20 #define HalF (l + r)>>1
 21 #define lsn rt<<1
 22 #define rsn rt<<1|1
 23 #define Lson lsn, l, mid
 24 #define Rson rsn, mid+1, r
 25 #define QL Lson, ql, qr
 26 #define QR Rson, ql, qr
 27 #define myself rt, l, r
 28 #define pii pair<int, int>
 29 #define MP(a, b) make_pair(a, b)
 30 using namespace std;
 31 typedef unsigned long long ull;
 32 typedef unsigned int uit;
 33 typedef long long ll;
 34 const int maxN = 2e5 + 7;
 35 int N, tot, root;
 36 struct Trie_node
 37 {
 38     int nex[26], fail; vector<int> val;
 39     Trie_node() { memset(nex, 0, sizeof(nex)); val.clear(); fail = 0; }
 40     void clear() { memset(nex, 0, sizeof(nex)); val.clear(); fail = 0; }
 41 } t[maxN];
 42 char s[maxN], T[2000006];
 43 void Insert(int ith)
 44 {
 45     int len = (int)strlen(s), u = root;
 46     for(int i=0, id; i<len; i++)
 47     {
 48         id = s[i] - 'a';
 49         if(!t[u].nex[id])
 50         {
 51             t[u].nex[id] = ++tot;
 52             t[tot].clear();
 53         }
 54         u = t[u].nex[id];
 55     }
 56     t[u].val.push_back(ith);
 57 }
 58 int que[maxN], top, tail;
 59 void build_fail()
 60 {
 61     top = tail = 0;
 62     que[tail++] = root;
 63     int tmp, p, son;
 64     while(top < tail)
 65     {
 66         tmp = que[top++];
 67         for(int i=0; i<26; i++)
 68         {
 69             son = t[tmp].nex[i];
 70             if(son)
 71             {
 72                 if(!tmp) t[son].fail = 0;
 73                 else
 74                 {
 75                     p = t[tmp].fail;
 76                     while(p && !t[p].nex[i]) p = t[p].fail;
 77                     t[son].fail = t[p].nex[i];
 78                 }
 79                 que[tail++] = son;
 80             }
 81             else t[tmp].nex[i] = t[t[tmp].fail].nex[i];
 82         }
 83     }
 84 }
 85 vector<int> to[maxN];
 86 int siz[maxN] = {0}, maxx;
 87 int ans[maxN];
 88 void dfs(int u)
 89 {
 90     for(int v : to[u])
 91     {
 92         dfs(v);
 93         siz[u] += siz[v];
 94     }
 95     if(!t[u].val.empty())
 96     {
 97         for(int i : t[u].val)
 98         {
 99             ans[i] = siz[u];
100         }
101     }
102 }
103 int main()
104 {
105     scanf("%d", &N);
106     tot = 0; root = 0;
107     t[root].clear();
108     for(int i=1; i<=N; i++)
109     {
110         scanf("%s", s);
111         Insert(i);
112     }
113     build_fail();
114     for(int i=0; i<=tot; i++) { to[i].clear(); siz[i] = 0; }
115     for(int i=1; i<=tot; i++) to[t[i].fail].push_back(i);
116     scanf("%s", T);
117     int u = root, len = (int)strlen(T);
118     for(int i=0, id; i<len; i++)
119     {
120         id = T[i] - 'a';
121         while(u && !t[u].nex[id]) u = t[u].fail;
122         u = t[u].nex[id];
123         siz[u]++;
124     }
125     maxx = 0;
126     dfs(root);
127     for(int i=1; i<=N; i++) printf("%d\n", ans[i]);
128     return 0;
129 }
posted on 2020-09-22 19:26 唔哩Wulili 阅读(175) 评论(0) 编辑收藏举报