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

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]

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]
" : "[Key No. %d] ", res[i]);
        }
    }
    else
    {
        printf( "No key can be found !
" );     
    }
    DeleteNode(pRoot);
    return 0;
}