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  • hdu 2846 Repository (字典树)

    Repository
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 7233    Accepted Submission(s): 2278

    Problem Description
    When you go shopping, you can search in repository for avalible merchandises by the computers and internet. First you give the search system a name about something, then the system responds with the results. Now you are given a lot merchandise names in repository and some queries, and required to simulate the process.
     
    Input
    There is only one case. First there is an integer P (1<=P<=10000)representing the number of the merchanidse names in the repository. The next P lines each contain a string (it's length isn't beyond 20,and all the letters are lowercase).Then there is an integer Q(1<=Q<=100000) representing the number of the queries. The next Q lines each contains a string(the same limitation as foregoing descriptions) as the searching condition.
     
    Output
    For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.
     
    Sample Input
    20
    ad
    ae
    af
    ag
    ah
    ai
    aj
    ak
    al
    ads
    add
    ade
    adf
    adg
    adh
    adi
    adj
    adk
    adl
    aes
    5
    b
    a
    d
    ad
    s
     
    Sample Output
    0
    20
    11
    11
    2

    C/C++:

     1 #include <map>
     2 #include <queue>
     3 #include <cmath>
     4 #include <vector>
     5 #include <string>
     6 #include <cstdio>
     7 #include <cstring>
     8 #include <climits>
     9 #include <iostream>
    10 #include <algorithm>
    11 #define INF 0x3f3f3f3f
    12 using namespace std;
    13 const int my_max = 1e6 + 10;
    14 
    15 struct node
    16 {
    17     int id, cnt, next[26];
    18 }my_node[my_max];
    19 int flag = 0;
    20 
    21 void my_init(int x)
    22 {
    23     for (int i = 0; i < 26; ++ i)
    24         my_node[x].next[i] = -1;
    25 }
    26 
    27 void my_insert(char *s, int len, int x)
    28 {
    29     int now = 0;
    30     for (int i = 0; i < len; ++ i)
    31     {
    32         if (my_node[now].next[s[i] - 'a'] == -1)
    33         {
    34             my_node[now].next[s[i] - 'a'] = ++ flag;
    35             now = flag;
    36             my_init(flag);
    37         }
    38         else now = my_node[now].next[s[i] - 'a'];
    39         if (my_node[now].id != x) my_node[now].cnt ++;
    40         my_node[now].id = x;
    41     }
    42 }
    43 
    44 int my_find(char *s)
    45 {
    46     int now = 0, len = strlen(s);
    47     for (int i = 0; i < len; ++ i)
    48     {
    49         if (my_node[now].next[s[i] - 'a'] == -1) return 0;
    50         now = my_node[now].next[s[i] - 'a'];
    51     }
    52     return my_node[now].cnt;
    53 }
    54 
    55 int main()
    56 {
    57     int n, len;
    58     scanf("%d", &n);
    59     char s[25];
    60     my_init(0);
    61     while (n --)
    62     {
    63         scanf("%s", s);
    64         len = strlen(s);
    65         for (int i = 0; i < len; ++ i)
    66             my_insert(s + i, len - i, n + 1);
    67     }
    68     scanf("%d", &n);
    69     while (n --)
    70     {
    71         scanf("%s", s);
    72         printf("%d
    ", my_find(s));
    73     }
    74     return 0;
    75 }
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  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9513845.html
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