HDU 1277 全文检索

全文检索

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1048 Accepted Submission(s): 324


[align=left]Problem Description[/align]
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。

[align=left]Input[/align]
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。

[align=left]Output[/align]
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 !。

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Found key: [Key No. 9] [Key No. 5]
题目大意：给定一段长数字串和一组短的数字串，问哪些短串在长串中出现过。
解题方法：AC自动机。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
using namespace std;

typedef struct node
{
int id;
node *fail;
node *next[10];
node()
{
id = 0;
fail = NULL;
memset(next, 0, sizeof(next));
}
}TreeNode;

int res[10005];
int nCount = 0;
bool flag = false;

void Insert(TreeNode *pRoot, char Substr[], int id)
{
int nLen = strlen(Substr);
TreeNode *p = pRoot;
for (int i = 0; i < nLen; i++)
{
int index = Substr[i] - '0';
if (p->next[index] == NULL)
{
p->next[index] = new TreeNode;
}
p = p->next[index];
}
p->id = id;
}

void BuildAC(TreeNode *pRoot)
{
queue<TreeNode*> Queue;
Queue.push(pRoot);
while(!Queue.empty())
{
TreeNode *p = Queue.front();
Queue.pop();
for (int i = 0; i < 10; 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 Query(TreeNode *pRoot, char str[])
{
TreeNode *p = pRoot;
int nLen = strlen(str);
for (int i = 0; i < nLen; i++)
{
int index = str[i] - '0';
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 && temp->id != -1)
{
if (temp->id > 0)
{
res[nCount++] = temp->id;
temp->id = -1;
flag = true;
}
temp = temp->fail;
}
}
}

void DeleteNode(TreeNode *pRoot)
{
for (int i = 0; i < 10; i++)
{
if (pRoot != NULL)
{
DeleteNode(pRoot->next[i]);
}
}
delete pRoot;
}

int main()
{
int m, n;
scanf("%d%d", &m, &n);
char temp[105];
char str[60001];
memset(str, 0, sizeof(str));
TreeNode *pRoot = new TreeNode;
for (int i = 0; i < m; i++)
{
scanf("%s", temp);
strcat(str, temp);
}
int num;
for (int i = 0; i < n; i++)
{
while(1)
{
char ch = getchar();
if (ch == ']')
{
getchar();
break;
}
}
scanf("%s", temp);
Insert(pRoot, temp, i + 1);
}
BuildAC(pRoot);
Query(pRoot, str);
if (flag)
{
printf("Found key: ");
for (int i = 0; i < nCount; i++)
{
printf(i == nCount - 1 ? "[Key No. %d]\n" : "[Key No. %d] ", res[i]);
}
}
else
{
printf( "No key can be found !\n" );
}
DeleteNode(pRoot);
return 0;
}