hdu 1277 AC自动机

全文检索

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1134    Accepted Submission(s): 357

Problem Description

我们大家经常用google检索信息，但是检索信息的程序是很困难编写的；现在请你编写一个简单的全文检索程序。
问题的描述是这样的：给定一个信息流文件，信息完全有数字组成，数字个数不超过60000个，但也不少于60个；再给定一个关键字集合，其中关键字个数不超过10000个，每个关键字的信息数字不超过60个，但也不少于5个；两个不同的关键字的前4个数字是不相同的；由于流文件太长，已经把它分成多行；请你编写一个程序检索出有那些关键字在文件中出现过。

 

Input

第一行是两个整数M，N；M表示数字信息的行数，N表示关键字的个数；接着是M行信息数字，然后是一个空行；再接着是N行关键字；每个关键字的形式是：[Key No. 1] 84336606737854833158。

 

Output

输出只有一行，如果检索到有关键字出现，则依次输出，但不能重复，中间有空格，形式如：Found key: [Key No. 9] [Key No. 5]；如果没找到，则输出形如：No key can be found !。

 

Sample Input

20 10
646371829920732613433350295911348731863560763634906583816269
637943246892596447991938395877747771811648872332524287543417
420073458038799863383943942530626367011418831418830378814827
679789991249141417051280978492595526784382732523080941390128
848936060512743730770176538411912533308591624872304820548423
057714962038959390276719431970894771269272915078424294911604
285668850536322870175463184619212279227080486085232196545993
274120348544992476883699966392847818898765000210113407285843
826588950728649155284642040381621412034311030525211673826615
398392584951483398200573382259746978916038978673319211750951
759887080899375947416778162964542298155439321112519055818097
642777682095251801728347934613082147096788006630252328830397
651057159088107635467760822355648170303701893489665828841446
069075452303785944262412169703756833446978261465128188378490
310770144518810438159567647733036073099159346768788307780542
503526691711872185060586699672220882332373316019934540754940
773329948050821544112511169610221737386427076709247489217919
035158663949436676762790541915664544880091332011868983231199
331629190771638894322709719381139120258155869538381417179544
000361739177065479939154438487026200359760114591903421347697

 

[Key No. 1] 934134543994403697353070375063

[Key No. 2] 261985859328131064098820791211

[Key No. 3] 306654944587896551585198958148

[Key No. 4] 338705582224622197932744664740

[Key No. 5] 619212279227080486085232196545

[Key No. 6] 333721611669515948347341113196

[Key No. 7] 558413268297940936497001402385

[Key No. 8] 212078302886403292548019629313

[Key No. 9] 877747771811648872332524287543

[Key No. 10] 488616113330539801137218227609

 

Sample Output

Found key: [Key No. 9] [Key No. 5]

 

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <algorithm>
 using namespace std;

 typedef struct node
 {
     int id;
     node *fail;
     node *next[];
     node()
     {
         id = ;
         fail = NULL;
         memset(next,NULL, sizeof(next));
     }
 }TreeNode;

 int res[];
 int nCount = ;

 void Insert(TreeNode *pRoot, char Substr[], int id)
 {
     int nLen = strlen(Substr);
     TreeNode *p = pRoot;
     for (int i = ; i < nLen; i++)
     {
         int index = Substr[i] - '';
         if (p->next[index] == NULL)
         {
             p->next[index] = new TreeNode;
         }
         p = p->next[index];
     }
     p->id = id;
 }

 void getFail(TreeNode *pRoot)
 {
     queue<TreeNode*> Queue;
     Queue.push(pRoot);
     while(!Queue.empty())
     {
         TreeNode *p = Queue.front();
         Queue.pop();
         for (int i = ; i < ; i++)
         {
             if (p->next[i] != NULL)
             {
                 if (p == pRoot)
                 {
                     p->next[i]->fail = pRoot;
                 }
                 else
                 {
                     TreeNode *temp = p->fail;
                     while(temp != NULL)
                     {
                         if (temp->next[i] != NULL)
                         {
                             p->next[i]->fail = temp->next[i];
                             break;
                         }
                         temp = temp->fail;
                     }
                     if (temp == NULL)
                     {
                         p->next[i]->fail = pRoot;
                     }
                 }
                 Queue.push(p->next[i]);
             }
         }
     }
 }

 void find(TreeNode *pRoot, char str[])
 {
     TreeNode *p = pRoot;
     int nLen = strlen(str);
     for (int i = ; i < nLen; i++)
     {
         int index = str[i] - '';
         while(p != pRoot && p->next[index] == NULL)
         {
             p = p->fail;
         }
         p = p->next[index];
         if (p == NULL)
         {
             p = pRoot;
         }
         TreeNode *temp = p;
         while(temp != pRoot)
         {
             if (temp->id > )
             {
                 res[nCount++] = temp->id;
                 temp->id = ;//同一个关键字只用一次
             }
             temp = temp->fail;
         }
     }
 }

 void DeleteNode(TreeNode *pRoot)
 {
     for (int i = ; i < ; i++)
     {
         if (pRoot != NULL)
         {
             DeleteNode(pRoot->next[i]);
         }
     }
     delete pRoot;
 }

 int main()
 {
     int i,m, n,c;
     char temp[];
     char str[];
     while(~scanf("%d%d", &m, &n))
     {
         strcpy(str,"");
         TreeNode *pRoot = new TreeNode;
         for (i = ; i < m; i++)
         {
             scanf("%s", temp);
             strcat(str, temp);
         }
         getchar();
         for (i = ; i < n; i++)
         {
             getchar();
             scanf("[Key No. %d] %s",&c,temp);
             Insert(pRoot, temp, i + );
         }
         getFail(pRoot);
         find(pRoot, str);
         if (nCount)
         {
             printf("Found key:");
             for (i = ; i < nCount; i++)
             {
                 printf(" [Key No. %d]", res[i]);
             }
             printf("\n");
         }
         else
         {
             printf( "No key can be found !\n" );
         }
     //    DeleteNode(pRoot);
     }
     return ;
 }