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  • hdu2222 Keywords Search ac自动机

    地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=2222

    题目:

    Keywords Search

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 56558    Accepted Submission(s): 18493


    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
    he
    say
    shr
    her
    yasherhs
     
    Sample Output
    3
     
    思路:ac自动机的模板题啊,我的ac自动机过的第一个题
      不过hdu的测评机好感人,第一交代码500msAC了,,后来由于某些原因再叫同一份代码时居然TLE了!!!再交时间从600-900多的都有。。。真是醉了
      1 #include <queue>
      2 #include <cstring>
      3 #include <cstdio>
      4 using namespace std;
      5 
      6 struct AC_auto
      7 {
      8     const static int LetterSize = 26;
      9     const static int TrieSize = 26 * ( 1e5 + 50);
     10 
     11     int tot,root,fail[TrieSize],end[TrieSize],next[TrieSize][LetterSize];
     12 
     13     int newnode(void)
     14     {
     15         memset(next[tot],-1,sizeof(next[tot]));
     16         end[tot] = 0;
     17         return tot++;
     18     }
     19 
     20     void init(void)
     21     {
     22         tot = 0;
     23         root = newnode();
     24     }
     25 
     26     int getidx(char x)
     27     {
     28         return x - 'a';
     29     }
     30 
     31     void insert(char *ss)
     32     {
     33         int len = strlen(ss);
     34         int now = root;
     35         for(int i = 0; i < len; i++)
     36         {
     37             int idx = getidx(ss[i]);
     38             if(next[now][idx] == -1)
     39                 next[now][idx] = newnode();
     40             now = next[now][idx];
     41         }
     42         end[now]++;
     43     }
     44 
     45     void build(void)
     46     {
     47         queue<int>Q;
     48         fail[root] = root;
     49         for(int i = 0; i < LetterSize; i++)
     50             if(next[root][i] == -1)
     51                 next[root][i] = root;
     52             else
     53                 fail[next[root][i]] = root,Q.push(next[root][i]);
     54         while(Q.size())
     55         {
     56             int now = Q.front();Q.pop();
     57             for(int i = 0; i < LetterSize; i++)
     58             if(next[now][i] == -1)   next[now][i] = next[fail[now]][i];
     59             else
     60                 fail[next[now][i]] = next[fail[now]][i],Q.push(next[now][i]);
     61         }
     62     }
     63 
     64     int match(char *ss)
     65     {
     66         int len,now,res;
     67         len = strlen(ss),now = root,res = 0;
     68         for(int i = 0; i < len; i++)
     69         {
     70             int idx = getidx(ss[i]);
     71             int tmp = now = next[now][idx];
     72             while(tmp)
     73             {
     74                 res += end[tmp];
     75                 end[tmp] = 0;//
     76                 tmp = fail[tmp];
     77             }
     78         }
     79         return res;
     80     }
     81     void debug(void)
     82     {
     83         for(int i = 0;i < tot; i++)
     84         {
     85             printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
     86             for(int j = 0;j < LetterSize;j++)
     87                 printf("%3d",next[i][j]);
     88             printf("]
    ");
     89         }
     90     }
     91 };
     92 
     93 AC_auto ac;
     94 char sa[1000001];
     95 int main(void)
     96 {
     97     int t,n;
     98     scanf("%d",&t);
     99     while(t--)
    100     {
    101         ac.init();
    102         scanf("%d",&n);
    103         while(n--)
    104             scanf("%s",sa),ac.insert(sa);
    105         ac.build();
    106         scanf("%s",sa);
    107         printf("%d
    ",ac.match(sa));
    108     }
    109 
    110     return 0;
    111 }
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  • 原文地址:https://www.cnblogs.com/weeping/p/5937136.html
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