HDU 3065 病毒侵袭持续中（AC自动机）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3065

题意：有n个模式串，一个主串，求每个模式串在主串中出现的次数

思路：AC自动机水题，trie树中也要维护标号，再开一个num数组记录出现次数即可，扫描匹配时注意跳转fail指针和root节点。

code：

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <string>

  4 #include <queue>

  5 using namespace std;

  6 const int KIND = 26;

  7 struct node

  8 {

  9     int id;

 10     node* fail;

 11     node* next[KIND];

 12     node ()

 13     {

 14         id = 0;

 15         fail = NULL;

 16         for (int i = 0; i < KIND; ++i) next[i] = NULL;

 17     }

 18 };

 19 node* root;

 20 int L;

 21 char str[1005][55];

 22 char text[2000005];

 23 int num[1005];

 24 

 25 void Insert(char str[])

 26 {

 27     node* temp = root;

 28     int len = strlen(str);

 29     for (int i = 0; i < len; ++i)

 30     {

 31         int curr = str[i] - 'A';

 32         if (temp->next[curr] == NULL)

 33             temp->next[curr] = new node();

 34         temp = temp->next[curr];

 35     }

 36     temp->id = ++L;

 37 }

 38 

 39 void Build()

 40 {

 41     queue<node*> Q;

 42     root->fail = root;

 43     for (int i = 0; i < KIND; ++i)

 44     {

 45         if (root->next[i] == NULL)

 46             root->next[i] = root;

 47         else

 48         {

 49             root->next[i]->fail = root;

 50             Q.push(root->next[i]);

 51         }

 52     }

 53     while (!Q.empty())

 54     {

 55         node* temp = Q.front();

 56         Q.pop();

 57         for (int i = 0; i < KIND; ++i)

 58         {

 59             if (temp->next[i] == NULL)

 60                 temp->next[i] = temp->fail->next[i];

 61             else

 62             {

 63                 temp->next[i]->fail = temp->fail->next[i];

 64                 Q.push(temp->next[i]);

 65             }

 66         }

 67     }

 68 }

 69 

 70 void Query()

 71 {

 72     node* temp = root;

 73     int len = strlen(text);

 74     for (int i = 0; i < len; ++i)

 75     {

 76         if (text[i] < 'A' || text[i] > 'Z')

 77         {

 78             temp = root;

 79             continue;

 80         }

 81         int curr = text[i] - 'A';

 82         while (temp->next[curr] == NULL && temp != root) temp = temp->fail;

 83         temp = temp->next[curr];

 84         if (temp == NULL) temp = root;

 85         node* x = temp;

 86         while (x != root)

 87         {

 88             if (x->id != 0) ++num[x->id];

 89             x = x->fail;

 90         }

 91     }

 92 }

 93 

 94 int main()

 95 {

 96     int n;

 97     while (scanf("%d", &n) != EOF)

 98     {

 99         L = 0;

100         root = new node();

101         for (int i = 1; i <= n; ++i)

102         {

103             scanf("%s", str[i]);

104             Insert(str[i]);

105         }

106         Build();

107         scanf("%s", text);

108         memset(num, 0, sizeof(num));

109         Query();

110         for (int i = 1; i <= n; ++i)

111         {

112             if (num[i] == 0) continue;

113             printf("%s: %d\n", str[i], num[i]);

114         }

115     }

116     return 0;

117 }

 数组实现：

  1 #include <cstdio>

  2 #include <cstring>

  3 #include <queue>

  4 using namespace std;

  5 const int KIND = 26;

  6 const int MAXN = 50005;

  7 const int MAXM = 2000005;

  8 

  9 struct Trie

 10 {

 11     int next[MAXN][KIND], fail[MAXN], id[MAXN], num[MAXN];

 12     int root, L, n;

 13     int create()

 14     {

 15         for (int i = 0; i < KIND; ++i)

 16             next[L][i] = -1;

 17         id[L++] = 0;

 18         return L - 1;

 19     }

 20     void init()

 21     {

 22         L = 0;

 23         n = 0;

 24         root = create();

 25         memset(num, 0, sizeof(num));

 26     }

 27     void insert(char str[])

 28     {

 29         int now = root;

 30         int len = strlen(str);

 31         for (int i = 0; i < len; ++i)

 32         {

 33             int curr = str[i] - 'A';

 34             if (-1 == next[now][curr])

 35                 next[now][curr] = create();

 36             now = next[now][curr];

 37         }

 38         id[now] = ++n;

 39     }

 40     void build()

 41     {

 42         queue<int>Q;

 43         fail[root] = root;

 44         for (int i = 0; i < KIND; ++i)

 45         {

 46             if (-1 == next[root][i])

 47                 next[root][i] = root;

 48             else

 49             {

 50                 fail[next[root][i]] = root;

 51                 Q.push(next[root][i]);

 52             }

 53         }

 54         while (!Q.empty())

 55         {

 56             int now = Q.front();

 57             Q.pop();

 58             for (int i = 0; i < KIND; ++i)

 59             {

 60                 if (-1 == next[now][i])

 61                     next[now][i] = next[fail[now]][i];

 62                 else

 63                 {

 64                     fail[next[now][i]] = next[fail[now]][i];

 65                     Q.push(next[now][i]);

 66                 }

 67             }

 68         }

 69     }

 70     void query(char str[])

 71     {

 72         int now = root;

 73         int len = strlen(str);

 74         for (int i = 0; i < len; ++i)

 75         {

 76             if (str[i] < 'A' || str[i] > 'Z')

 77             {

 78                 now = root;

 79                 continue;

 80             }

 81             now = next[now][str[i] - 'A'];

 82             int temp = now;

 83             while (temp != root)

 84             {

 85                 if (id[temp]) ++num[id[temp]];

 86                 temp = fail[temp];

 87             }

 88         }

 89     }

 90 };

 91 Trie ac;

 92 char str[1005][55];

 93 char text[MAXM];

 94 int main()

 95 {

 96     int n;

 97     while (scanf("%d", &n) != EOF)

 98     {

 99         ac.init();

100         for (int i = 1; i <= n; ++i)

101         {

102             scanf("%s", str[i]);

103             ac.insert(str[i]);

104         }

105         ac.build();

106         scanf("%s", text);

107         ac.query(text);

108         for (int i = 1; i <= n; ++i)

109         {

110             if (0 == ac.num[i]) continue;

111             printf("%s: %d\n", str[i], ac.num[i]);

112         }

113     }

114     return 0;

115 }