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统计难题 HDOJ--2222

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29287 Accepted Submission(s): 9572

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

Input

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

Output

Print how many keywords are contained in the description.

Sample Input

1 5

she

say

shr

her

yasherhs

Sample Output

思路： AC自动机

AC代码：

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<stdlib.h>
  4 typedef struct Node_Tree
  5 {
  6     int cnt; 
  7     struct Node_Tree *child[26]; 
  8     struct Node_Tree *fail; 
  9 }Node; 
 10 Node *root; 
 11 char keywd[51]; 
 12 char decpt[1000001]; 
 13 Node *q[500001]; 
 14 int tail = 0, head = 0; 
 15 void insert()
 16 {
 17     if(keywd == NULL)
 18         return ; 
 19     int i; 
 20     char *p = keywd;
 21     Node *t = root;
 22     while(*p != '')
 23     {
 24         if(t->child[*p - 'a'] == NULL)
 25         {
 26             Node *temp = (Node *)malloc(sizeof(Node)); 
 27             memset(temp, 0, sizeof(Node)); 
 28             for(i = 0; i < 26; i ++)
 29             {
 30                 temp->child[i] = NULL; 
 31             }
 32             temp->cnt = 0;
 33             temp->fail = NULL; 
 34             t->child[*p - 'a'] = temp; 
 35         }
 36         t = t->child[*p - 'a']; 
 37         p ++; 
 38     }
 39     t->cnt ++; 
 40 }
 41 
 42 void getfail()
 43 {
 44     int i; 
 45     q[tail++] = root;
 46     while(tail != head)             //BFS; 
 47     {
 48         Node *p = q[head++];
 49         Node *temp = NULL;
 50         for(i = 0; i < 26; i ++)
 51         {
 52             if(p->child[i] != NULL)
 53             {
 54                 if(p == root)
 55                 {
 56                     p->child[i]->fail = root; 
 57                 }
 58                 else
 59                 {
 60                     temp = p->fail; 
 61                     while(temp != NULL)
 62                     {
 63                         if(temp->child[i] != NULL)
 64                         {
 65                             p->child[i]->fail = temp->child[i]; 
 66                             break ; 
 67                         }
 68                         temp = temp->fail; 
 69                     }
 70                     if(temp == NULL)
 71                         p->child[i]->fail = root; 
 72                 }
 73                 q[tail++] = p->child[i]; 
 74             }
 75         }
 76     }
 77 }
 78 
 79 int search()
 80 {
 81     int i, ret = 0; 
 82     char *p = decpt; 
 83     Node *t = root;
 84     while(*p != '')
 85     {
 86         while(t->child[*p - 'a'] == NULL && t != root)
 87             t = t->fail; 
 88         t = t->child[*p - 'a']; 
 89         if(t == NULL)
 90             t = root;
 91         Node *temp = t; 
 92         while(temp != root && temp->cnt != -1)
 93         {
 94             ret += temp->cnt;
 95             temp->cnt = -1;
 96             temp = temp->fail;
 97         }
 98         p ++; 
 99     }
100     return ret; 
101 }
102 
103 int main(int argc, char const *argv[]) 
104 {
105     int c, i, t; 
106     scanf("%d", &c);
107     Node TREEROOT; 
108     root = &TREEROOT; 
109     while(c --)
110     {
111         for(i = 0; i < 30; i ++)
112         {
113             root->child[i] = NULL; 
114             root->cnt = 0;
115             root->fail = NULL; 
116         }
117         tail = head = 0; 
118         memset(decpt, 0, sizeof(decpt));
119         memset(keywd, 0, sizeof(keywd)); 
120         scanf("%d", &t); 
121         while(t --)
122         {
123             scanf("%s", keywd); 
124             insert(); 
125             memset(keywd, 0, sizeof(keywd)); 
126         }
127         getfail(); 
128         scanf("%s", decpt);
129         printf("%d
", search());
130     }
131     return 0; 
132 }

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原文地址：https://www.cnblogs.com/anhuizhiye/p/3484845.html